ALGEBRAIC TOPOLOGY Preamble. This Document Contains Some Exercises in Algebraic Topology, Category Theory, and Homological Algeb

Home , Boundary (topology), Cofibration, Inclusion map, Mapping cone (topology), Puppe sequence, Unit interval

KLINT QINAMI

Preamble. This document contains some exercises in algebraic topology, category theory, and homological algebra. Most of them can be found as chapter exercises in Hatcher’s book on algebraic topology. Use at your own risk.

Exercise 1. If a topological space X is contractible, then it is path-connected.

Proof. Let F be the homotopy between the identity map and the constant map fpxq “ x0 for some contraction point x0 P X. For all x P X, let γx “ F |txuˆI . γx is continuous since restrictions of continuous functions are continuous. γxp0q “ F px, 0q “ idpxq “ x and γxp1q “ F px, 1q “ fpxq “ x0, hence it is a well-deﬁned path from x to x0. Thus, for any 1 x1, x2 P X, there exists a path from x1 to x2 given by γx2 ˚ γx1 . Exercise 2. If f, f 1 : X Y are homotopic and g, g1 : Y Z are homotopic, then Ñ x Ñ g ˝ f and g1 ˝ f 1 are homotopic.

Proof. Let F and G denote the homotopies between f, f 1 and g, g1 respectively. Let Hpx, tq “ GpF px, tq, tq. H is continuous and Hpx, 0q “ GpF px, 0q, 0q “ Gpfpxq, 0q “ gpfpxqq and Hpx, 1q “ GpF px, 1q, 1q “ Gpf 1pxq, 1q “ g1pf 1pxqq, hence g ˝ f and g1 ˝ f 1 are homotopic.

Exercise 3. (a) The composition of homotopy equivalences X Ñ Y and Y Ñ Z is a homotopy equivalence X Ñ Z. Homotopy equivalence is an equivalence relation. (b) The relation of homotopy among maps X Ñ Y is an equivalence relation. (c) A map homotopic to a homotopy equivalence is a homotopy equivalence.

Proof. paq Let f : X Ñ Y and g : Y Ñ X be continuous functions such that f ˝g is homotopic to idY and g ˝ f is homotopic to idX . Similarly, let h, k be continuous functions such that h ˝ k „ idZ and k ˝ h „ idY . Composing functions we have pg ˝ kq ˝ ph ˝ fq “ g ˝ pk ˝ hq ˝ f „ g ˝ idY ˝ f „ g ˝ f „ idX and ph ˝ fq ˝ pg ˝ kq “ h ˝ pf ˝ gq ˝ k „ h ˝ idY ˝ k „ h ˝ k „ idZ , where we have used associativity of function composition and that homotopy is well-behaved with respect to function composition (c). Since homotopy equivalence is reﬂective, symmetric, and transitive, it is an equivalence relation. pbq For any map f, f is homotopic to itself by the homotopy that is f identically throughout the unit interval. If f „ g by F px, tq, then g „ f by F px, 1´tq. Lastly, homotopy is transitive since homotopies can be pasted together by doubling speeds and gluing. Hence, homotopy among functions is an equivalence relation. pcq It suﬃces to show that if f „ f 1, then f ˝ g „ f 1 ˝ g. But if F px, tq is the homotopy from f to f 1, then F ˝ g ˆ id is the homotopy from f ˝ g to f 1 ˝ g.

1A wide hat over a path denotes the inverse path. That is, γpsq “ γp1 ´ sq 1 y Exercise 4. (a) A space X is contractible iﬀ every map f : X Ñ Y , for arbitrary Y, is null-homotopic. (b) Similarly, X is contractible iﬀ every map f : Y Ñ X is null-homotopic.

Proof. paq p ùñ q Let F be a contraction of X to some contraction point x0. Then for any map f : X Ñ Y , we have f ˝ F : X ˆ I Ñ Y giving the homotopy between f and the constant map x ÞÑ fpx0q. p ðù q If every map f : X Ñ Y for arbitrary Y is null-homotopic, then in particular idX is null-homotopic. pbq p ùñ q Let F be a contraction of X to some contraction point x0. Then for any map f : Y Ñ X, F pfpyq, tq is the desired homotopy from f to y ÞÑ x0. p ðù q If every map f : Y Ñ X for arbitrary Y is null-homotopic, then in particular idX is null-homotopic. Exercise 5. (a) f : X Ñ Y is a homotopy equivalence if there exist maps g, h : Y Ñ X such that fg – 1 and hf – 1. (b) More generally, f is a homotopy equivalence if fg and hf are homotopy equivalences.

Proof. paq f is a homotopy equivalence since fphfgq “ fphfqg – fp1qg – fg – 1 and phfgqf “ hpfgqf – hp1qf – 1. pbq This is entirely similar to the proof of paq.

Exercise 6. For a path-connected space X, π1pXq is abelian iﬀ all basepoint-change homomorphisms βh depend only on the endpoints of the path h.

1 Proof. p ùñ q Consider any two paths h, h from x0 to x1. Consider any loop γ based ´1 1 1´1 at x0. We have βhrγs “ rhγh s and βh1 rγs “ rh γh s. Conjugating again, we have ´1 1´1 1 1´1 1 ´1 rh sβhrγsrhs “ rγs and rh sβh1 rγsrh s “ rγs. Hence, βhrγs “ rhh sβh1 rγsrh h s “ 1´1 1 ´1 rhh srh h sβh1 rγs “ βh1 rγs, since π1pXq is abelian. p ðù q Let γ1 and γ2 be two loops based at x0. Decompose and reparametrize γ1 into 1 1 1 two paths δ1 “ γpr0, sq and δ2 “ γpr , 1sq. By assumption, we have that βδ1 rγ2s “ β ´ rγ2s. 2 2 δ2 ´1 ´1 Explicitly, this gives rδ1γ2δ1 s “ rδ2 γ2δ2s. Multiplying on the right by δ1 and on the left by δ2 gives rδ2δ1γ2s “ rγ2δ2δ1s. But δ2δ1 “ γ1, so π1pXq is abelian. Exercise 7. For a space X, the following three conditions are equivalent: (a) Every map S1 Ñ X is homotopic to a constant map, with image a point. (b) Every map S1 Ñ X 2 extends to a map D Ñ X. (c) π1pX, x0q “ 0 for all x0 P X. (d) It then follows that a space X is simply-connected iﬀ all maps S1 Ñ X are homotopic. [Here, âĂŸhomotopicâĂŹ means âĂŸhomotopic without regard to basepointsâĂŹ.]

Proof. pa ùñ bq Let f : S1 Ñ X be any map and let h denote a homotopy from a constant map to f. Then the extension of f is just given by the homotopy, fpθ, rq “ hpθ, rq, where θ, r give the usual angle-radius parametrization of the disk. For r “ 1, we have fpθ, 1q “ hpθ, 1q “ fpθq. r pb ùñ cq Let x0 be any point in X. Given an equivalence class in π1pX, x0q, a representa- tiver γ is a map S1 Ñ X, so it extends to a map D2 Ñ X, but the map it extends to is exactly a based homotopy to a constant loop. Hence, every loop based at x0 is null-homotopic, so π1pX, x0q is trivial. 1 pc ùñ aq The hypothesis π1pX, x0q for all x0 P X implies that all maps S Ñ X homotopic to the trivial loop, and hence homotopic to a constant map. 2 pdq p ùñ q If a space is simply connected, then π1pXq “ 0, so pcq holds, and thus paq holds. But since X is path-connected, all maps homotopic to a constant map are homotopic. p ðù q If all maps S1 Ñ X are homotopic, then in particular, the constant maps are homotopic, and hence X is path-connected. Additionally, paq holds, which implies π1pXq “ 0.

Exercise 8. From the isomorphism π1pX ˆY, px0, y0qq – π1pX, x0qˆπ1pY, y0q, it follows that loops in X ˆty0u and tx0uˆY represent commuting elements of π1pX ˆY, px0, y0qq. The following is an explicit homotopy demonstrating this.

Proof. Let γx : I Ñ X ˆ ty0u and γy : I Ñ tx0u ˆ Y be loops based at px0, y0q. Let

x0 0 ď 2t ď s δxpt, sq “ $π1pγxp2t ´ sqq s ď 2t ď 1 ` s &’x0 1 ` s ď 2t ď 2

%’y0 0 ď 2t ď 1 ´ s δypt, sq “ $π2pγyp2t ´ sqq 1 ´ s ď 2t ď 2 ´ s &’x0 2 ´ s ď 2t ď 2 Let Hpt, sq “ pδ pt, sq, δ pt, sqq. H is continuous as δ and δ are continuous. Hpt, 0q “ x y %’ x y pδxpt, 0q, δypt, 0qq “ γxγy and Hpt, 1q “ γyγx and Hp0, sq “ Hp1, sq “ px0, y0q, so H is a based homotopy and rγxγys “ rγyγxs.

Exercise 9. For a covering map p : X Ñ X and a subspace A Ă X, let A “ p´1pAq. The restriction p : A Ñ A is a covering map. r r Proof. Since p is a covering map, there exists an open cover tUαu of X by evenly covered r ´1 i sets Uα. That is, for all α, p pUαq “ iPI Vα for some index set I, and the restriction i p : Vα Ñ Uα for any particular i is a homeomorphism. Since the Uα cover X, tUα X Au is an ´Ů1 ´1 ´1 i ´1 open cover of A. Additionally, for all α, p pUα XAq “ p pUαqXp pAq “ iPI Vα Xp pAq. i i ´1 i ´1 p : Vα Ñ Uα is a homeomorphism, so the restriction p : Vα X p pAq Ñ ppVα X p pAqq is also i ´1 Ů a homeomorphism. But ppVα X p pAqq “ Uα X A since a homeomorphism is in particular bijective, so the Uα X A are evenly covered. Hence A has an open cover by evenly covered 1 sets, so the restriction p : p´ pAq Ñ A is a covering map.

Exercise 10. Let X and Y be simply-connected covering spaces of the path-connected, locally path-connected spaces X and Y . If X » Y , then X » Y . r r Proof. Let p : X X and q : Y Y be covering maps and let f : X Y and g : Y X Ñ Ñ r r Ñ Ñ be homotopy equivalences such that fg – idY and gf – idX . Since p is a covering map, pfpq˚pπ1pXqq correspondsr to ther trivial subgroup, and X and Y are path-connected and locally-path connected, so by the lifting criterion, fp, and similarly gq, extend to lifts fp : X Ñ Y andr gq : Y Ñ X such that qfp “ fp and pgq “ gq. Since pgqfp “ gqfp “ gfp, by the lifting lemma applied to the homotopy gf –Ă1, therer existsr ar homotopyr r pgqfp to p.Ă This homotopyr also lifts to a homotopy from gqfp to p : X Ñ Xr.,Ă where p Ăis a lift of the covering map p. But p is a deck transformation,r Ă since pp “ p and p is a homeomorphism. r Ă 3 r r r r r r r ´1 ´1 Hence, p gqfp – p p “ idX . A similar construction gives idY , and hence by Proposi- tion 5, X and Y are homotopy equivalent under fp and gq. r r r r Ă r r Lemma 1. Let p : X Ñ Y be a covering map and let Y be locally path-connected. Then X is locallyr path-connected.r Ă r Proof. Consider any point x P X and an open neighborhood U of x. Let V denote an evenly covered neighborhood of ppxq. Let W be an open neighborhood of x such that p : W Ñ V is a homeomorphism. U X W is then homeomorphic to ppU X W q, which contains ppxq. But Y is path-connected, so there exists an open neighborhood O Ă ppU X W q that is path- connected. Then the inverse image of O under the local homeomorphism is a path-connected open neighborhood of x that is a subset of U, so X is locally path-connected.

Exercise 11. For a covering map p : X Ñ X with X connected, locally path-connected, and semi-locally simply-connected, (a) the components of X are in one-to-one correspon- ´1 dence with the orbits of the action of π1rpX, x0q on the fiber p px0q. (b) Under the Galois correspondence between connected covering spaces of X andr subgroups of π1pX, x0q, the subgroup corresponding to the component of X containing a given lift x0 of x0 is the stabilizer of x0, the subgroup consisting of elements whose action on the fibers leaves x0 fixed. r r r r Proof. paq By Lemma 1, each connected component of X is locally path-connected, and ´1 hence path-connected. Consider any two elements x1 and x2 of the fiber p px0q in the same component of X. Since the component is path-connected,r there exists a path γ from x1 to x2. But then rpγs P π1pX, x0q, and prpγsq ¨ px1q “ xr2, hencer x1 and x2 are in the same orbit. ´1 Now conversely,r suppose x1 and x2 are points in the fiber p px0q and are in the same orbit.r Explicitly,r there exists rγs P π1pX, x0q such thatr rγrs ¨x1 “ x2.r But thenr by the lifting lemma, γ lifts to a path from x1 tor x2, sor x1 and x2 must be in the same component. pbq Let x0 be a given lift of x0. The stabilizer of xr0 is ther set of all loops classes rγs such that rγs¨x0 “ x0. Butr then anyr suchr γ mustr lift to a loop based at x0. These are precisely the elements inr π1pX, x0q. Each statement is reversible,r so the double containment holds. r r r 1 1 Exercise 12.r Definer f : S ˆ I Ñ S ˆ I by fpθ, sq “ pθ ` 2πs, sq, so f restricts to the identity on the two boundary circles of S1 ˆ I. f is homotopic to the identity by a homotopy ft that is stationary on one of the boundary circles, but not by any homotopy ft that is stationary on both boundary circles.

Proof.r ft is given explicitly by ftpθ, sq “ pθ ` 2πts, sq. Suppose for a contradiction that a homotopy ft from f to the identity ﬁxing both boundary circles existed. Then ft gives 1 a based homotopy from the trivial loop to the generator of π1pS q, pπ ˝ f0 ˝ iqpsq “ 0 to 2 pπ ˝ f1 ˝ iqpsqr “ 2πs, which cannot exist. r r 1 1 Exerciser 13. Every homomorphism π1pS q Ñ π1pS q can be realized as the induced 1 1 homomorphism φ˚ of a map φ : S Ñ S .

2Here i denotes the inclusion map, and π has been overloaded as the natural projection map, the ratio of the circumference of a circle to its diameter, and as the fundamental group when sub-scripted by 1. 4 Proof. Let ψ : π1pSq Ñ π1pSq be any homomorphism. ψ is determined by the image of iθ ikθ 2πs the generator, ψprγ1sq “ rγks, since π1pSq – Z. Let φ : e ÞÑ e . If γ1psq “ e , then φ ˝ γ1 “ γk, so φ˚ “ ψ.

Exercise 14. There are no retractions r : X Ñ A in the following cases: (a) X “ R3 with A any subspace homeomorphic to S1. (b) X “ S1 ˆ D2 with A its boundary torus S1 ˆ S1. (c) X “ S1 ˆ D2 and A the circle shown in the ﬁgure.

(d) X “ D2 _ D2 with A its boundary S1 _ S1. (e) X a disk with two points on its boundary identiﬁed and A its boundary S1 _ S1. (f) X the MÃűbius band and A its boundary circle.

3 Proof. paq If such a retract existed, there would be a surjective homomorphism π1pR q – 1 1 Ñ π1pS q – Z. 1 pbq If such a retract existed, there would be a surjective homomorphism from π1pS ˆ 2 1 1 D q “– Z ˆ t0u – Z Ñ π1pS ˆ S q – Z ˆ Z. 1 pcq The generator of π1pAq maps to a loop in S that is homotopic to the trivial loop, since it laps around and then backwards, so the homomorphism cannot be surjective, and hence no retract exists. 2 2 pdq If such a retract existed, we would have a surjective homomorphism from π1pD _D q – 1 1 1 Ñ π1pS _ S q – Z ˚ Z. peq X deformation retracts onto S1, so if such a retract existed to S1 _ S1, there would 1 1 1 be a surjective homomorphism from π1pS q – Z Ñ π1pS _ S q – Z ˆ Z. pfq X deformation retracts to its central circle. Let γ be a loop around A. Composing this loop with a loop going around the central circle of X gives a loop that goes around the central circle twice. Hence the inclusion of A in X induces a homomorphism from Z Ñ 2Z that restricts to the identity on 2Z, which is impossible, so no such retract exists.

Exercise 15. If a path-connected, locally path-connected space X has π1pXq ﬁnite, then every map X Ñ S1 is null-homotopic.

1 Proof. Let f : X Ñ S be any map. Since π1pXq is ﬁnite, f˚pπ1pXqq is a ﬁnite subgroup of Z and hence trivial. By the Lifting Criterion, there exists a lift f : X Ñ R of f. R is contractible, so f is null-homotopic by some homotopy h. But if p : R Ñ S1 is a covering map, then p ˝ h is a homotopy from f to a constant map. r r Exercise 16. For a path-connected, locally path-connected, and semilocally simply- connected space X, call a path-connected covering space X abelian if it is normal and has an abelian deck transformation group. X has an abelian covering space that is a covering space of every other abelian covering space of X, and suchr a âĂŸuniversalâĂŹ abelian covering space is unique up to isomorphism. Below is a description of this covering space explicitly for X “ S1 _ S1 and X “ S1 _ S1 _ S1. 5 Proof. The commutator subgroup rπ1pXq, π1pXqs is a normal subgroup, and hence corresponds to a normal, path-connected covering space X. The group of deck transformations of X is also abelian, since the quotient of a group by the commutator subgroup yields that groups abelianization. r Supposer there exists another covering map q : X Ñ X with q˚pπ1pXqq normal and the quotient π1pXq{q˚pπ1pXqq abelian. The commutator subgroup of π1pXq lies inside q˚pπ1pXqq, so by the lifting criterion, the map p : X Ñ X liftsp to a map p : X Ñ Xp . But the lift of a covering map is a coveringp map, so X covers X. p Now suppose actually that X is alsor a universal abelian cover.r r Thenp both p and q are lifts of the covering maps satisfying rpqp “ p,p so by the uniqueness of lifts, qp must be the identity on X. A similar argumentp shows that pq is the identity on X, so we mustr havep an isomorphism between X and X. pr pr 1 1 Concretely,r if X “ S _ S , then we are lookingrp for a space whose groupp of deck transformations is the abelianizationr ofp Z ˚ Z, which is Z ˆ Z. This would be some two dimensional lattice where horizontal movements correspond to one generator and vertical movements correspond to another generator. The case for S1 ˆ S1 ˆ S1 should be a similar but three dimensional lattice. Exercise 17. Given a covering space action of a group G on a path-connected, locally path-connected space X, then each subgroup H ă G determines a composition of covering spaces X Ñ X{H Ñ X{G. Then (a) Every path-connected covering space between X and X{G is isomorphic to X{H for some subgroup H ă G. (b) Two such covering spaces X{H1 and X{H2 of X{G are isomorphic if and only if H1 and H2 are conjugate subgroups of G. (c) The covering space X{H Ñ X{G is normal if and only if H is a normal subgroup of G, in which case the group of deck transformations of this cover is G{H.

p1 p2 Proof. paq Given a sequence X ÝÑ Y ÝÑ X{G, let H “ tg P G | p1 ˝ g “ p1u. p1 descends to a map p1 : X{H Ñ Y since p1 is constant on equivalence classes. That is, if x1 „H x2, then hpx1q “ x2 for some h P H. But then p1px1q “ p1hpx1q “ p1px2q. Define a map q : Y Ñ X{H by qpyq “ rxsH , where x is an element in the fiber of y under p1. This map is well-defined r 1 1 1 since if p1pxq “ p1px q “ y, then p2p1pxq “ p2p1px q, so x „G x . But then p1 ˝ gpxq “ p1pxq, so p1 ˝ g “ p1, and hence g P H. So Y and X{H are isomorphic since p1 is a continuous bijection preserving the covering. pbq p ùñ q Given an isomorphism f : X{H1 Ñ X{H2. Since f is an isomorphism,r we have qf “ p where p and q are the covering maps from X{H1 Ñ X{G and X{H2 Ñ X{G. If 1 1 rxsH1 ÞÑ rx sH2 under f, then there exists g such that gx “ x since both qf “ p. ´1 p ðù q Let g be such that gH1g “ H2. Define the map f : rxsH1 ÞÑ rgxsH2 . This map is well defined since if hpxq “ x1, then there exists h1 such that h1gx “ gx1, namely h1 “ ghg´1. The map is a bijection since its inverse is given by the map generated by g´1. Given an open set U in X{H2, can take the inverse image under the natural projection to get an open set ´1 in X. But the taking the inverse isomorphism g and projecting back onto X{H1 gives an 1 open set since the projection map is open, equivalent to f ´ pUq.

6 Exercise 18. The complement of a finite set of points in Rn is simply connected if n ě 3. n Proof. Let S “ tx1, x2, . . . , xmu be a finite set of points in R . Let 2 “ mini‰j |xi ´ xj| for i P N such that 0 ă i ă m ` 1. Connect Bpxiq to Bpxi`1q by a path for every i P N such that 0 ă i ă m. Then the complement of S deformation retracts to the boundaries of each ball together with the paths between them (first deformation retract onto a ball containing the entire system, then deflate the ball). But this is homotopy equivalent to m n´1 n´1 m n´1 m n´1 m i“1 S . Each S is a CW complex, so π1p i“1 S q “ ˚i“1 π1pS q “ ˚i“1 0 “ 0 by Van-Kampen. Ž Ž 3 3 Exercise 19. Let X Ă R be the union of n lines through the origin. π1pR ´ Xq “ Z˚p2n´1q. Proof. R3 ´ X deformation retracts onto S2 minus 2n points, where the 2n points are the intersections of S2 with the lines and the deformation retract is the usual one onto the unit sphere along rays from the origin. But by the stereographic projection, S2 ´ pS2 X Xq is homeomorphic to R2 minus 2n ´ 1 points. R2 minus 2n ´ 1 points has a bouquet of 2n ´ 1 2n 1 circles as a deformation retract, so its fundamental group is Z˚p ´ q.

1 Exercise 20. The fundamental group obtained from two tori by identifying S ˆ tx0u in 1 one torus to S ˆ tx0u in the other torus is pZ ˚ Zq ˆ Z. Proof. The identiﬁcation space is

This gives a presentation for the group as G “ xa, b, c abb´1a´1 “ e, bcc´1b´1 “ ey. We have the free group on three generators, where one of the generators commutes with both of the others. This gives pZ ˚ Zq ˆ Z. The space is just pS1 _ S1q ˆ S1. To see this, place one torus inside the hole of the other, and identify the inner circle of the outer torus with the outer circle of the inner torus. This 1 1 1 1 1 1 1 1 1 gives π1ppS _S qˆS q “ π1pS _S qˆπ1pS q “ pπ1pS q˚π1pS qqˆπ1pS q “ pZ˚ZqˆZ.

2 2 Exercise 21. π1pR ´ Q q is uncountable. 2 2 2 2 Proof. Consider px0, x0q P R ´ Q . For any px, xq P R ´ Q with x0 ă x, let γx be the box path from px0, x0q Ñ px, x0q Ñ px, xq Ñ px0, xq Ñ px0, x0q. For rational r such that x0 ă r ă x, pr, rq is enclosed by γx, so γx is not trivial. It remains to show γx is not 1 homotopic to γx1 for x ‰ x . Without loss of generality, assume x ă x1. Again there exists rational r with x ă r ă x1. 2 2 2 Consider the inclusion map i : pR ´ Q q ãÑ pR ´ tpr, rquq. rγxs is in the kernel of i˚, but γx1 isn’t, so γx and γx1 are not homotopic. Exercise 22. Of the following, only pdq is not a category.

7 (a) Objects are ﬁnite sets, morphisms are injective maps of sets. (b) Objects are sets, morphisms are surjective maps of sets. (c) Objects are abelian groups, morphisms are isomorphisms of abelian groups. (d) Objects are sets, morphisms are maps of sets which are not surjective. (e) Objects are topological spaces, morphisms are homeomorphisms. Proof. paq Composition of injective maps is injective. Composition is associative. The identity is an injection. pbq Composition of surjective maps is surjective. Composition is associative. The identity is a surjection. pcq Composition of isomorphisms is an isomorphism and composition is associative. The identity is an isomorphism. pdq The identity map is surjective, so the identity is not a morphism, so this can’t be a category. peq Composition of homeomorphisms is a homeomorphism and composition is associative. The identity is a homeomorphism. Exercise 23. Below are examples of categories with (a) One object and four morphisms. (b) Two objects and ﬁve morphisms.

Proof. paq We can regard Z4 as a category with one object and four morphisms. The object is the underlying set, and the morphisms are the maps given by adding by an element.

A B pbq

Exercise 24. The simplest possible category is the empty category 0, consisting of no objects and no morphisms. Given another category C, there is a unique functor F : 0 Ñ C, taking nothing nowhere. By definition, colim F is called an initial object of C, if it exists. (a) Any two initial objects in a category are uniquely isomorphic. (b) Below is a description of which of the following categories have initial objects, and what they are: Set, Gp, Top, Top*, the category of fields with field homomorphisms, the category of infinite-dimensional vector spaces over a given field with linear maps, the category of small categories with functors Cat. (c) Below is a definition of the notion of a terminal object in a category, and a description of the terminal objects (if they exist) in the previous categories.

Proof. paq Suppose 0 and 01 are initial objects in C. Then there exist unique morphisms f : 0 Ñ 01 and g : 01 Ñ 0. Composing, we have fg and gf as unique morphisms 01 Ñ 01 and 0 Ñ 0. But the respective identity morphisms are such morphisms, so they must be the identity morphisms by uniqueness. Hence 0 and 01 are isomorphic, and the isomorphism is unique. pbq An initial object is an object such that for every object in the category, there exists precisely one morphism from the initial object to that object. 8 Set has the empty set as its initial object, since theres only one map from the empty set to any other set. Gp has the trivial group as its initial object since there’s only one homomorphism from the trivial group to any other group, sending the identity to the identity. Top has the empty set as its initial object since theres only one function from the empty set to any other topological space.

Top˚ has the one point space ˚ as its initial object, since the point must go to the base point of any other space under any continuous map. Field has no initial object. Vec8 has no initial object. Cat has the empty category as its initial object since theres only one map from the empty category to any other category taking nothing nowhere. pcq A terminal object is an object such that for every object in the category, there is a unique morphism from that object to the terminal object. Set has any one element set as a terminal object, where the only maps to the one element set are the maps sending everything to that set. Gp has the trivial group as its terminal object as well, since the only map to the trivial group sends everything to the identity. Top has ˚ as a terminal object since any map to ˚ sends everything to ˚. Top˚ has the one point space ˚ as its terminal object since any map sends everything to ˚. Field has no terminal object. Vec8 has no terminal object. Cat has the category with one object and one morphism as its terminal object, since any functor must send everything to that one object. Exercise 25. A natural isomorphism is a natural transformation α, say between two functors F,G : C Ñ D, such that αX is an isomorphism for each X P ObpCq. Two categories C and D are equivalent if there exist functors F : C Ñ D and G : D Ñ C such that there are natural isomorphisms : FG Ñ idD and η : idC Ñ GF . Let X be a topological space and x P X. Regard π1pX, xq, a group, as a category with one element x. Let Π1pXq be the fundamental groupoid of X: its objects are points of X and its morphisms from x to y are homotopy classes of paths (with ﬁxed endpoint) from x to y. There is an evident âĂĲinclusion functorâĂİ J : π1pX, xq Ñ Π1pXq. If X is path-connected, J is an equivalence of categories.

Proof. Following the proof by Peter May, define the inverse functor F : Π1pXq Ñ π1pX, xq where F pAq is an object isomorphic to A in π1pX, xq. choosing an isomorphism αA : A Ñ ´1 F pAq and mapping morphisms f : A Ñ B as F pfq “ αB ˝f ˝αA : F pAq Ñ F pBq. Let αA be the identity if A P π1pX, xq. Then FJ “ id and αA : id Ñ JF is a natural isomorphism. Exercise 26. Given two maps of topological spaces f : X Ñ Z and g : Y Ñ Z, let X ˆZ Y , the pullback or fiber product of f and g, be defined as the set

X ˆZ Y “ tpx, yq P X ˆ Y fpxq “ gpyqu

9 equipped with the subspace topology inherited from the product, together with the âĂĲob- viousâĂİ maps X ˆZ Y Ñ X and X ˆZ Y Ñ Y formed by composing inclusion and projection onto one of the factors. The pullback is a limit in Top over the diagram X Ñ Z Ð Y . Proof. Let T, x, y be a cone over the same diagram. It suﬃces to show there exists a morphism u from T to the pullback such that the following diagram commutes.

x T u p y X ˆZ Y X q f g Y Z

Let uptq “ pxptq, yptqq. This is well-deﬁned since gpyptqq “ fpxptqq, so uptq P X ˆZ Y for all t P T . u is continuous since the coordinate functions are continuous and the pullback has the subspace topology inherited from the product topology. More explicitly,the product topology has rectangular open sets as a basis. The preimage of any basis element under u is the intersection of the preimages under x and y, which is open as it is the intersection of two open sets. Exercise 27. Given two maps of topological spaces f : Z Ñ X and g : Z Ñ Y , the pushout of f and g is deﬁned as the quotient space X Y “ X Y {„ Z ž ž where „ is the equivalence relation generated by fpzq „ gpzq for all z P Z, together with

the “obvious” maps X Ñ X Z Y and Y Ñ X Z Y formed by composing inclusion into X Y and the quotient map. The pushout is a colimit in Top over the diagram X Ð Z Ñ Y . š š š Proof. Let T, j1, j2 be a cocone over the same diagram. It suﬃces to show there exists a morphism u from the pushout to T such that the following diagram commutes.

T j2 u X Y Y Z i2 j1 g iš1 X Z f

j1pwq w P i1pXq Let upwq “ . Both j1 and j2 are continuous and agree on i1pXqXi2pY q as #j2pwq w P i2pY q j1 ˝f “ j2 ˝g. Both i1pXq and i2pY q are closed, so by the gluing lemma, u is continuous.

10 Exercise 28. The mapping cylinder Mf of a map f : X Ñ Y is a colimit over the diagram X ˆ I Ð X Ñ Y , where X Ñ X ˆ I is the natural map identifying X with X ˆ t0u.

Proof. Mf is the pushout X ˆ I X Y with the maps f : X Ñ Y and i : X Ñ X ˆ I identifying X to X ˆ t0u, and hence by the previous proposition, it is a colimit over the š diagram X ˆ I Ð X Ñ Y . Exercise 29. All CW complexes with two 0-cells and two 1-cells up to (a) homeomorphism are in one of the four classes described below. (b) homotopy equivalence are in one of the three classes described below.

1 1 Proof. paq We have two maps f : ta, bu Ñ tx1, x2u and g : ta , b u Ñ tx1, x2u, giving 16 possible constructions. But up to relabeling of nodes and changing directions of edges (homeomorphism), we only have

x1 x2 x1 x2 x1 x2 x1 x2

pbq The second box from the left is homotopy equivalent to the last box from the left by a contraction of the x2 node to the x1 node.

Exercise 30. (a) The ‘square lattice’ is a CW complex homeomorphic to R2. (b) The following diagram is a CW complex homeomorphic to a 2-disk with two smaller open 2-disks removed.

Proof. paq Let the 0-skeleton be ZˆZ. To construct the 1-skeleton, connect nodes a distance one apart to one-another by the path of distance one. To construct the 2-skeleton, ﬁll in each square pi, jq Ñ pi ` 1, jq Ñ pi ` 1, j ` 1q Ñ pi, j ` 1q Ñ pi, jq. pbq The one skeleton is given by 11 x1

Gluing a disk to the large left region and a disk to the large right region gives the desired disk with two smaller open disks removed.

Exercise 31. RP n ´txu, where x P RP n is any point, is homotopy equivalent to RP n´1.

Proof. RP n is Sn{pv „ ´vq, but this is equivalent to Dn with antipodes on BDn identiﬁed. BDn with antipodes identiﬁed is just RP n´1, so the real projective space of dimension n can n´1 n n´1 n´1 be constructed by R f D where the attaching map f : S Ñ RP is the quotient projection. If the point to be removed lies on the boundary of Dn, use a homeomorphism to move it to the interior.Ť Then Dn minus an interior point deformation retracts onto its boundary, so RP n minus a point deformation retracts to RP n´1, and hence they are homotopy equivalent.

Exercise 32. (a) The mapping cylinder of every map f : S1 Ñ S1 is a CW complex. (b) The follwoing is a 2 dimensional CW complex that contains both an annulus S1 ˆI and a MÃűbius band as deformation retracts.

Proof. paq Mf in this instance can be constructed explicitly as a CW complex. Let the 0-skeleton be two points, θ0 and fpθ0q. Attach a 1-cell to θ0 and a 1-cell to fpθ0q, forming a loop at each point. Attach a 1-cell connecting θ0 to fpθ0q. Attach a 2-cell along the path going along the loop at θ0, along the 1-cell joining θ0 to fpθ0q, along the image of the loop at s0 under f, then back along the path joining the 0-cells. pbq Place CW structures on the annulus and the MÃűbius band, and identify the central circle of the MÃűbius band with S1 ˆt0u. But then this CW complex retracts onto both the annulus and the MÃűbius band, by using the retraction of the MÃűbius band to its central 1 circle and the retract of the annulus to S ˆ t0u, respectively.

12 Exercise 33. Below is a description of which of the following inclusions are cofibrations. (a) txu ãÑ Sn, where x P Sn is any point. (b) p0, 1s ãÑ r0, 1s. (c) Z ãÑ R. (d) Q ãÑ R. (e) X ãÑ CX, where CX “ pX ˆ Iq{pX ˆ t0uq is the cone of X and the inclusion sends X Ñ X ˆ t1u. Proof. For the following, consider the diagram Y X f0 p0 i r f Y I A f r A cofibration exists if and only if there is a retraction X ˆ I to A ˆ I Y X ˆ t0u. But if A Ă X, then the cofibration, if it exists, must be the inclusion, since a cofibration is injective and is a homeomorphism onto its image. paq This is a cofibration since there exists a retraction from Sn ˆ I to pI ˆ txu Y pX ˆ t0uq. pbq A cofibration cannot exist since there is no retraction from I2 to pp0, 1sÎqYpI ˆt0uq. pcq This is a cofibration since there is a retraction from R ˆ I to pZ ˆ Iq Y pR ˆ t0uq. pdq A cofibration cannot exist since there is nor etraction from RÎ to pQÎqYpRˆt0uq. peq This is a cofibration since there exists a retract from CXÎ to pXÎqYpCXˆt0uq. Exercise 34. If f : X Ñ Y is a (co)fibration, and g : Y Ñ Z is a (co)fibration, then g ˝ f : X Ñ Z is a (co)fibration. Proof. Let W be any topological space, h : W ˆ I Ñ Z be any continuous map, i be the map sending W to W ˆ t0u, and h0 be such that h ˝ i “ g ˝ f ˝ h0. g ˝ f is a fibration iff there exists a function u such that Diagram A commutes. Well, g is a fibration, so using f ˝ h0, there exists a lift h such thatr Diagram B commutes. f is anotherr fibration, so there must exist a lift h such that h ˝ i “ h0. Then letting u “ h gives commutative Diagram C, so gr˝ f is a fibration. p r r r r Diagram A Diagram B Diagram C W h0 X W h0 X W h0 X

r f r f r f u u h i Y i Y i Y r g h g h g W ˆ I h Z W ˆ I ph Z W ˆ I ph Z

Reversing arrows in the diagrams gives a proof that coﬁbrations are closed under composition. 13 Exercise 35. If ˚ is the one point space, then any map f : X Ñ ˚ is a ﬁbration.

Proof. Let Z be any topological space, i be the map sending Z to Z ˆ t0u, and let h and h0 be continuous maps such that h ˝ i “ f ˝ h0. Then f is a ﬁbration iﬀ there exists a map u such that the following diagram commutes r r Z h0 X ur i f Z ˆ I h ˚

Let upz, tq “ h0pzq. Then u ˝ i “ h0, and f ˝ u “ h since any map X Ñ ˚ is constant, so the diagram commutes. r r I Exercise 36. A map p : E Ñ B is a ﬁbration iﬀ the map π : E Ñ Ep, πpγq “ pγp0q, pγq, I has a section, that is, a map s : Ep Ñ E such that πs “ 1.

Proof. p ùñ q Consider the following diagram

pe,γqÞÑe Ep E s p Epˆt0u

Ep ˆ I B pe,γ,tqÞÑγptq The diagram commutes since γp0q “ ppeq, so the lift s exists since p is a ﬁbration. We have πpspe, γqq “ pspe, γqp0q, pspe, γqq. But by the diagram, spe, γqp0q “ e and pspe, γq “ γ, so πs “ 1. p ðù q It suﬃces to show there exists f making

f Xr 0 E fr i p r f X ˆ I B commute. Since πs “ 1, πpspe, γqq “ pspe, γqp0q, pspe, γqq “ pe, γq so spe, γqp0q “ e and

ppsqpe, γq “ γ. Let fpx, tq “ pspf0pxq, ftpxqqqptq. This is a valid lift since fpx, tq “ pspf0pxq, f0pxqqqp0q “ f0pxq as spe, γqp0q “ e. It is continuous since s, f0 and ft are continuous. The diagram commutesr sincer pf “ ppspf0pxq, ftpxqqptqq “ fpx, tq as ppsqpe,r γq “ γ. r r r r r Exercise 37. A linear projection of a 2-simplex onto one of its edges is a ﬁbration but not a ﬁber bundle.

Proof. Let T be any 2 simplex in R2. Project onto a side where both incident angles are less than π{2, which must exist since otherwise the angles would sum to more than π. Then the fiber of an endpoint belonging to that edge is a single point, while the fiber of any point that’s not an endpoint is a line, so the fibers are not homeomorphic, and thus the projection is not a fiber bundle. 14 I It remains to show the projection is a fibration. We have π : T Ñ Tp as in the previous proposition given by πpγq “ pγp0q, pγq where p : T Ñ Te is the projection onto the edge. It suffices to construct a section such that πs “ 1. The section is given explicitly by using a path that is mimics the path on the edge, except it slides along edges if the path it tries to mimic goes outside the triangle. Exercise 38. If X and Y are pointed topological spaces, then xΣX,Y y “ xX, ΩY y. Proof. Let φ : F pX, ΩY q Ñ F pΣX,Y q be given by φphqpx, sq “ hpxqpsq. φ is well-defined since φphqpx, s0q “ hpxqps0q “ y0 and φphqpx0, sq “ hpx0qpsq “ y0, where y0 denotes the loop 1 ´1 ´1 staying put at y0 for all s P S . Let φ pgqpxqpsq “ gpx, sq. Then we have φpφ pgqqpx, sq “ φ´1pgqpxqpsq “ gpx, sq and φ´1pφphqqpxqpsq “ φphqpx, sq “ hpx, sq. Hence φ is a bijection. But φ is a restriction of the usual currying function, and since S1 is locally compact Hausdorff, so the currying function is continuous, and in particular φ. The inverse is continuous since it is the un-currying function. It remains to show that φ descends on based homotopy equivalence classes. But if H is a homotopy between h, h1 P F pX, ΩY q, then φ ˝ H is a homotopy between φh and φh1. The composition is a valid homotopy since φ and H are continuous. Exercise 39. Suppose that X is a CW complex that is an increasing union of subcomplexes X1 Ă X2 Ă ... such that each inclusion Xj ãÑ Xj`1 is nullhomotopic. Then X is contractible.

Proof. Each pair pX,Xjq satisfies the homotopy extension property since the Xj are subcomplexes, and thus each null-homotopy hj : Xj ˆ I Ñ Xj`1 extends to a homotopy hj : X ˆ I Ñ X. As in the proof of inclusion of subcomplexes being cofibrations, define h : X ˆ I Ñ X to be the composition of all of the homotopy extensions. This gives a rcontraction of X.

Exercise 40. Given a pointed space pX, x0q, view X as X ˆ I{pX ˆ t0u Y X ˆ t1u Y tx0u ˆ Iq. Then the inclusion i : X ãÑ X given by x ÞÑ px, 1{2q is nullhomotopic. ř 1 1 Proof. Let hpx, tq “ px, 2 p1 ` tqq. Then hřpx, 0q “ px, 2 q and hpx, 1q “ px, 1q. But under the equivalence, all points px, 1q are identified to a single point. Exercise 41. If X is a pointed CW complex, then the infinite suspension Σ8pXq “ ΣjpXq jě1 ď with inclusions as in the previous proposition is contractible. Since Σ8 is a functor, this gives a way of making arbitrary CW complexes contractible. In particular, S8 is contractible. Proof. Σ8pXq is an increasing union of subcomplexes X Ă ΣX Ă Σ2X Ă ... . Each inclusion i : ΣjX ãÑ Σj`1 is null-homotopic by Proposition 7, so by Proposition 6, the infinite suspension is contractible. Exercise 42. If a diagram

15 f g A B C h D i E

j l k m n

1 o 1 p 1 q 1 r 1 Ar Br C Dr Er of homomorphisms of abelian groups with exact rows commutes, and all vertical maps except the middle are isomorphisms, then the middle map is injective. Proof. Let 0 denote the identity element in a group. Let c be an element in the kernel of k. Any homomorphism maps 0 ÞÑ 0, so qpkpcqq “ 0. Since the diagram commutes, qpkpcqq “ mphpcqq “ 0. But m is an isomorphism, so hpcq “ 0. By exactness of the ﬁrst row, there exists an element b P B such that gpbq “ c. Since the diagram commutes, pplpbqq “ kpgpbqq “ rkpcq “ 0. By exactnessr of the second row, there exists a1 such that opa1q “ lpbq. j is an isomorphism, so there exists a P A such that jpaq “ a1. The diagram commutes,r so lfpaqq “ opjpaqq “ opa1q “ lpbq. Since j is injective, fpaq “ b. But gpfpaqq “ gpbq “ c rand rgpfpaqq “ 0 by exactness of the diagram, so ker k “ 0r and hence k is injective. r Assumptions:r maps are homomorphisms,r r rows are exact, commutativity of the diagram, and j and m are isomorphisms.

Exerciser r 43. If p : pE, e0q Ñ pB, b0q is a based ﬁbration of based spaces, then the ´1 inclusion φ : F “ p pbq Ñ Np is a based homotopy equivalence.

I Proof. Explicitly, Np “ tpe, γq P E ˆ B : γp0q “ ppeq, γp1q “ b0u. Let h : Np ˆ I Ñ B be given by htpe, γq “ γptq. Let h0 : Np Ñ E be given by h0pe, γq “ e. By the homotopy lifting property, there exists a lift h : Np ˆ I Ñ E such that r r

h0 r Np E hr i p r h Np ˆ I B

commutes. Let H : Np ˆ I Ñ Np be given by Htpe, γq “ phtpe, γq, γ|rt,1sq. Note that Ht is ﬁber preserving since the endpoints of the paths are unchanged. 1 The map e ÞÑ pe, cppeqq is a homeomorphism between E andr its image in Np, call it E . Note that H1pe, γq “ ph1pe, γq, cγp1qq, and since by commutativity of the diagram, pph1pe, γqq “ 1 h1pγq “ γp1q, H1pNpq Ă E – E. We can thus regard H1 as a map Np Ñ E. Then φH1 “ H1. 1 Additionally, we haver that HtpEq Ă E – E for all t. r Note that H0 “ id, so that by Ht, we have that φH1 is based homotopy equivalent to the identity. Additionally, H1φ is based homotopy equivalent to the identity by Ht|E, so φ is a based homotopy equivalence of ﬁbers.

Exercise 44. Let f : X Ñ Y be a map of pointed topological spaces and let π : Nf Ñ X be the projection. Then π is always a ﬁbration, and the natural inclusion i : ΩY Ñ Nπ is a homotopy equivalence.

Proof. Let 16 h0 Z Nf hr i π r Z ˆ I h X

be a commutative diagram without h. π is a ﬁbration if and only if there exists h such that the diagram still commutes.

If such a lift exists, since π ˝ h “ hr, the first coordinate of the lift must be htpzrq1 “ htpzq. To be well defined, the second coordinate of the lift must be a path from fphtpzqq to y0, where y0 is the basepoint of Y .r Interpret and reparametrize htpzq as a pathr from h0pzq to htpzq. Then, a path from fphtpzqq to fph0pzqq is given by γzptq “ fphtpzqq, where htpzq denotes the inverse path to htpzq. Concatenate this path with the second coordinate of the initial lift h0pzq2. This is well-defined since h0pzq2 must be a path frompfph0pzq1q to yp0 be definition of Nf , and by commutativity of the diagram, h0pzq “ πph0pzqq, so h0pzq “ h0pzq1. Hence ther lift h exists as is given explicitly byr htpzq “ phtpzq, h0pzq2 ˚ γzptqqr. r r n n Exercise 45.r For n P N Y t8u, πipS q – πiprRP q for all irą 1.

n n n n Proof. Pick arbitrary basepoints s0 P S and p0 P RP . Let π : pS , s0q Ñ pRP , p0q be the basepoint preserving natural projection identifying antipodal points. π is a covering map by 0 ´1 Proposition 1.40 in Hatcher, so therefore it is also a ﬁbration. Identify S as π ptp0uq 1 n with an arbitrary basepoint s0. Let i : S0 Ñ S be the basepoint preserving inclusion. Then we have the long exact sequence of homotopy groups 0 1 n n 0 1 ... Ñ πipS , s0q Ñ πipS , s0q Ñ πipRP , p0q Ñ πi´1pS , s0q Ñ ... 0 1 which follows from the long exact sequence of a ﬁbration. For i ą 1, πipS , s0q is trivial, so the short exact sequence

f n g n h 0 ÝÑ πipS , s0q ÝÑ πipRP , p0q ÝÑ 0 is valid for all i ą 1. g is injective since ker g “ fpt0uq “ 0, and it is surjective since its image n n is ker h. Therefore, πipS , s0q – πipRP , p0q. The proposition follows since both spaces are path-connected. This proposition holds also for S8 and RP 8 since S8 is a double cover of RP 8.

m n n m Exercise 46. Let m, n P Zą1 Y t8u and let X “ RP ˆ S and Y “ RP ˆ S (pick basepoints arbitrarily). Then πipXq – πipY q for all i ě 0.

Proof. Real projective spaces and spheres of dimension greater than one are path-connected, m n m n n m so by Proposition 4.2 in Hatcher, πipRP ˆ S q – πipRP q ˆ πipS q and πipRP ˆ S q “ n m πipRP q ˆ πipS q. The case i “ 0 follows since both spaces are path-connected. n m n By 1B.3 in Hatcher, π1pRP q – Z{2Z for n P Zą1Yt8u. Hence, for i “ 1, π1pRP ˆS q – m n n m n m π1pRP qˆπ1pS q – Z{2Zˆ0 – Z{2Z and π1pRP ˆS q “ π1pRP qˆπipS q – Z{2Zˆ0 – Z{2Z. The case i ą 1 follows by the previous proposition.

17 Exercise 47. Let

gn`1 in fn gn in´1 ... Ñ Cn`1 ÝÝÝÑ An ÝÑ Bn ÝÑ Cn ÝÑ An´1 ÝÝÑ Bn´1 Ñ ...

be a long exact sequence of abelian groups, with every third arrow in injective. Then the short sequence g in fn h 0 ÝÑ An ÝÑ Bn ÝÑ Cn ÝÑ 0 is exact.

Proof. in is an injective homomorphism, so ker in “ 0 “ gpt0uq. im in “ ker fn since the long exact sequence is exact. Additionally, the exactness of the long sequence gives im gn “ ker in´1, which is trivial since in´1 is injective. Then im gn “ 0 gives ker gn “ Cn. By exactness of the long sequence, im fn “ ker gn “ Cn. Hence im f “ Cn “ ker h, so the short sequence is exact. Exercise 48. If f j 0 ÝÑe A ÝÑ B ÝÑh C ÝÑ 0 is a short exact sequence of abelian groups, and there exists a map g : B Ñ A such that gf “ idA, then B – A ‘ C. Proof. Let φ : B Ñ A ‘ C be given by φpbq “ pgpbq, hpbqq. φ is a homomorphism since its coordinate functions are homomorphisms. Suppose b P ker φ. Then gpbq “ 0 and hpbq “ 0. b P ker h, hence there exists a P A such that fpaq “ b by exactness of the sequence. This gives that gpfpaqq “ gpbq “ 0. But gf “ idA, so gpfpaqq “ b, hence ker φ is trivial. Therefore, φ is injective. Consider any pa, cq P A ‘ C. h is surjective by exactness, so there exists b such that hpbq “ c. Let b1 “ fpaq ` b ´ fpgpbqq. Then φpb1q “ pgpfpaq ` b ´ fpgpbqqq, hpfpaq ` b ´ fpgpbqqqq “ pgpfpaqq ` gpbq ´ gpfpgpbqqq, hpfpaqq ` hpbq ´ hpfpgpbqqqq g, h are homomorphisms

“ pa ` gpbq ´ gpbq, hpfpaqq ` hpbq ´ hpfpgpbqqqq gf “ idA “ pa, hpbqq “ pa, cq im f “ ker h φ is a bijective homomorphism, so it is an isomorphism, and therefore B – A ‘ C. Exercise 49. If pX,Aq is a pointed pair such that there exists a retraction r : X Ñ A, then πipXq – πipAq ‘ πipX,Aq for all i ě 2. Proof. Consider the relative homotopy sequence of pairs derived from the Puppe Sequence given by

... Ñ πipAq Ñ πipXq Ñ πipX,Aq Ñ πi´1pAq Ñ ...

Since there is a retraction X Ñ A, the map πipAq Ñ πipXq is injective for all i ě 2. By Proposition 6, the short sequence

0 Ñ πipAq Ñ πipXq Ñ πipX,Aq Ñ 0 is exact for all i ě 2. By deﬁnition of retract, the retract composed with the inclusion is the identity map on A. But then by functoriality, the composition of the induced maps is the identity on the homotopy groups. That is, pr ˝ iq “ id implies pr ˝ iq˚ “ r˚ ˝ i˚ “ id˚. Thus, 18 by Proposition 7, the short exact sequence splits, and hence πipXq – πipAq ‘ πipX,Aq for all i ě 2. Exercise 50. An n-dimensional, n-connected CW complex is contractible. Proof. Let X be an n-dimensional, n-connected CW complex. By CW approximation, there exists ΓX such that γ : X Ñ ΓX is a weak-homotopy equivalence and ΓX has a unique 0-cell and no q-cells for 0 ă q ď n. By Whitehead’s Theorem, this weak-equivalence is a homotopy equivalence. By cellular approximation, γ is homotopy equivalent to a cellular map. But then this map is a homotopy equivalence from X to a point, since X has no q-cells for q ą n and each q-cell for q ď n gets mapped to the unique 0-cell of ΓX. Hence X is contractible. Exercise 51. A CW complex retracts onto any contractible subcomplex.

Proof. Let ht : A Ñ A be a homotopy such that h0paq “ a0 for some a0 P A and let h1paq “ a. Let H0 : X Ñ A be given by H0pxq “ a0. Note H0|A “ h0. Since any CW pair satisﬁes the homotopy extension property, there exists an extension Ht : X Ñ A such that Ht|A “ ht. But then H1 gives the desired retract, since H1 : X Ñ A and H1|A “ h1 “ idA. Lemma 2. Let Z be a CW approximation of a space X. If there exists a weak homotopy equivalence X Ñ Y or Y Ñ X, then Z is a CW approximation of Y . Proof. Consider ﬁrst the case where there exists a weak homotopy equivalence g : X Ñ Y . Let h : Z Ñ X be the given weak homotopy equivalence. Then g ˝ h : Z Ñ Y is a weak homotopy equivalence from Z to Y , so Z is a CW approximation for Y . In the latter case, there exists a weak homotopy equivalence f : Y Ñ X. Let W be a CW approximation of Y and let w : W Ñ Y be the corresponding weak homotopy equivalence. By the previous argument using composition, W is a CW approximation of X. But then by Corollary 4.19 in Hatcher, there is a homotopy equivalence k : Z Ñ W . Hence, w ˝ k : Z Ñ Y is a weak homotopy equivalence, hence Z is a CW approximation for Y .

Exercise 52. Consider the equivalence relation »w generated by weak homotopy equivalence: X »w Y if there are spaces X “ X1,X2,...,Xn “ Y with weak homotopy equivalences Xi Ñ Xi`1 or Xi Ð Xi`1 for each i. X »w Y iﬀ X and Y have a common CW approximation. Proof. ( ùñ ) Let Z be a CW approximation for X. Then by n-fold application of Lemma 1, Z is a CW approximation for Y , and hence they have a common CW approximation. ( ðù ) If Z is a CW approximation of X and Y , then we have the sequence X Ð Z Ñ Y where both arrows are weak homotopy equivalences since Z is a CW approximation. But then this is the criterion for X »w Y .

Exercise 53. There is no retraction RP n Ñ RP k if n ą k ą 0. Proof. Suppose there is a retraction r : RP n Ñ RP k for n ą k ą 0. Then there must be a n k n k surjection πkpRP q Ñ πkpRP q. If k ą 1, then πkpRP q “ 0 and πkpRP q “ Z. But this n gives a contradiction since there is no surjection 0 Ñ Z. If k “ 1, then πkpRP q “ Z{2Z k and πkpRP q “ Z, and again there is a contradiction since there is no surjection Z{2Z Ñ Z. Thus no retraction exists. 19 Exercise 54. Given a sequence of CW complexes KpGn, nq, n “ 1, 2,..., let Xn be the CW complex formed by the product of the first n of these KpGn, nq’s. Via the inclusions Xn´1 Ă Xn coming from regarding Xn´1 as the subcomplex of Xn with the n-th coordinate equal to a basepoint 0-cell of KpGn, nq, we can then form the union of all the Xn’s, a CW complex X. πnpXq – Gn for all n. Proof. The projection map X to a factor is a fiber bundle with canonical sections, and hence we have the long sequence of homotopy groups. But because of the section, each short exact sequence splits to give that πnpXq – Gn ˆ 0 ˆ 0 ... – Gn. Exercise 55. For a fiber bundle F Ñ E Ñ B such that the inclusion F ãÑ E is homotopic to a constant map, the long exact sequence of homotopy groups breaks up into split short exact sequences giving isomoprhisms πnpBq – πnpEq‘πn´1pF q. In particular, for the Hopf bundles S3 Ñ S7 Ñ S4 and S7 Ñ S15 Ñ S8, this yields isomorphisms 4 7 3 πnpS q – πnpS q ‘ πn´1pS q 8 15 7 πnpS q – πnpS q ‘ πn´1pS q 4 8 Thus π7pS q and π15pS q have Z summands.

Proof. Since F ãÑ E is homotopic to a constant map, the induced map πnpF q Ñ πnpEq is trivial for all n. By exactness, the kernel of the map πnpEq Ñ πnpBq is trivial, and thus the map is injective. Therefore, the sequence

0 Ñ πnpEq Ñ πnpBq Ñ πn´1pF q Ñ 0

is exact. To see that the sequence splits, we construct a map πi´1pF q Ñ πipBq. Since F ãÑ E is homotopic to a constant map, each map Si´1 Ñ F bounds a disk Di Ñ E. Composing this disk map with the projection gives a map which sends to boundary of the i disk to a single point, so it represents a map S Ñ B, and gives an element of πipBq. Since this map is a well-defined homomorphism, the short exact sequence splits by and πnpBq – πnpEq ‘ πn´1pF q. Exercise 56. If Sk Ñ Sm Ñ Sn is a fiber bundle, then k “ n ´ 1 and m “ 2n ´ 1. Proof. For each point p P Sn, there exists an open neighborhood U such that U ˆ Sk – π´1pUq. U ˆ Sk is an open subset of Sn ˆ Sk, and π´1pUq is an open subset of Sm, so by Theorem 2.26 in Hatcher, m “ n ` k. We proceed by cases on n. If n “ 0, then m “ k by m “ n ` k, and the fiber bundle has the form Sk Ñ Sk Ñ S0. The map Sk Ñ S0 has to be surjective, and thus Sk must be disconnected, so k “ 0. But then the fiber bundle is S0 Ñ S0 Ñ S0, which is impossible, since a surjective map S0 Ñ S0 must be a bijection as the spaces have the same finite cardinality, and hence cannot have fiber S0. If n “ 1, then the fiber bundle has the form Sk´1 Ñ Sk Ñ S1 by m “ n ` k. The exact k 1 k´1 sub-sequence π1pS q Ñ π1pS q Ñ π0pS q implies k “ 1 since otherwise there would exist an exact sequence 0 Ñ Z Ñ 0. So then the fiber bundle is of the form S0 Ñ S1 Ñ S1, which agrees with k “ n ´ 1 and m “ 2n ´ 1. m If n ą 1 then k ă m since m “ n ` k, and hence πkpS q “ 0. Therefore any map k m n m k S Ñ S is null-homotopic, and by Proposition 6, πnpS q – πnpS q ‘ πn´1pS q. But 20 m n k k n ă m, so πnpS q “ 0 and hence πnpS q – πn´1pS q – Z. In particular, πn´1pS q is non- n k m trivial, so n ´ 1 ě k. Additionally, by the exactness of πk`1pS q Ñ πkpS q Ñ πk´1pS q – 0, n k n πk`1pS q Ñ πkpS q – Z is surjective, so πk`1pS q cannot be trivial, and thus k ě n ´ 1. Thus, k “ n ´ 1. Finally, since m “ n ` k, m “ 2n ´ 1. Exercise 57. Let X be the triangular parachute obtained from ∆2 by identifying its three 2 vertices to a single point. Its homology groups are H0pXq – Z,H1pXq – Z , with higher homology groups being trivial.

Proof. Since the space is path-connected, we have that H0pXq – Z. H1pXq “ kerpB1q{ impB2q. Since all vertices are identified, B1 is the trivial map, and hence its kernel is xa, b, cy. We still have impB2q “ á ` b ´ c, and hence kerpB1q{ impB2q “ xa, b, cy{xá ` b ´ cy – Z2. Since impB3q “ kerpB2q “ 0, we have that H2pXq – 0. The higher homology groups vanish since the boundary maps for all higher simplices are trivial. Exercise 58. Let X be the ∆ complex X obtained from ∆n by identifying all faces of the same dimension. Thus X has a single k-simplex for each k ď n. The homology groups of X all vanish except H0pXq – Z and HnpXq if n is even.

Proof. The case H0pXq follows since the space is path-connected. For 0 ă i ă n, we have

i j θi´1 i even Ci{Ci – 0 i even Biθi “ j“0p´1q θ|i-face “ . Hence, HipXq – . Lastly, #0 i odd #0 i odd HnpXqř “ kerpBnq{ impBn`1q – kerpBnq. Therefore, HnpXq – Z if n is even and trivial otherwise.

Exercise 59. If A is a retract of X, then the map HnpAq Ñ HnpXq induced by the inclusion is injective.

Proof. Let x P ker i˚. Then pr ˝ iq˚pxq “ pr˚ ˝ i˚qpxq “ 0. Hence, ker i˚ Ă kerpr ˝ iq˚ “ ker id˚ “ 0, so ker i˚ “ 0, and hence i˚ is injective. Exercise 60. A morphism f : A Ñ B in Ch(Ab) is an isomorphism if and only if each fn : An Ñ Bn is an isomorphism of groups.

Proof. ( ùñ ) If f is an isomorphism, then f|An is an isomorphism onto its image Bn since the restriction of the inverse of f is the inverse of the restriction of f. ( ðù ) We have the following diagram

Bi Bi`1 ... Ai Ai`1 ...

fi fi`1 1 1 Bi Bi`1 ... Bi Bi`1 ...

Let gi be the inverse of fi. It suﬃces to show that the collection of all gi is a morphism. But 1 this follows since fi`1Bi “Bifi by deﬁnition of morphism, and by composition on the left by 1 gi`1 and on the right by gi, we have Bigi “ gi`1Bi, and hence the inverse is a morphism so f is an isomorphism. 21 Exercise 61. If A is a complex with all boundary maps equal to the zero map, then HnpAq – An.

Proof. We have HnpAq “ kerpBnq{ impBn`1q – An{0 – An. Exercise 62. Given two morphisms f, g : A Ñ B in Ch(Ab), let the sum f `g : A Ñ B be given by pf ` gqn “ fn ` gn. Hn : Ch(Ab) Ñ Ab is additive: Hnpf ` gq “ Hnpfq ` Hnpgq.

Proof. Hnpf ` gq “ kerppf ` gqnq{ imppf ` gqn`1q “ kerpfn ` gnq{ impfn`1 ` gn`1q “ Hnpfq ` Hnpgq.

1 1 Exercise 63. A is a subcomplex of a complex A if each An is a subgroup of An and each boundary map on A1 is the restriction of the corresponding boundary map on A. If f : A Ñ B is a morphism of complexes, then there is an isomorphism of complexes A{ kerpfq – impfq.

Proof. By the ﬁrst isomorphism theorem on groups, we have An{ kerpfnq – impfnq. But the isomoprhism is the quotient map of f, so Bg “ gB, and hence by Proposition 4, we have that there is an isomorphism in Ch(Ab) of A{ kerpfq – impfq.

2 1 1 Exercise 64. (a) Compute the homology groups HnpX,Aq when X is S or S ˆS and A is a ﬁnite set of points in X. (b) Compute the groups HnpX,Aq and HnpX,Bq for X a closed, orientable surface of genus two with A and B the circles shown. [ What are X/A and X/B. ]

Proof. paq A is a ﬁnite subset of a metric space, hence closed in particular. If A is empty, then the relative homology groups HnpX,Aq reduce to the homology groups HnpXq, which have already been computed in Hatcher. Otherwise, A is nonempty with cardinality k P N. Let U be the open set given by unions of disjoint open balls centered at each point in A. Then U deformation retracts onto A by contracting each ball to its center. Thus pX,Aq is a good pair. k A has k-path components, so H0pAq “ Z . Furthermore, A contains no maps of higher dimensional simplices, hence HipAq – 0 for i ą 0. X is path-connected in both cases, so X{A is path-connected. Thus, H0pX,Aq – H0pX{Aq – 0. Consider the long exact sequence

... Ñ H1pAq Ñ H1pXq Ñ H1pX,Aqr Ñ H0pAq Ñ H0pXq Ñ H0pX,Aq Ñ 0 2 2 If X “ S , then H1pS q – 0, so we have the short exact sequence k 0 Ñ H1pX,Aq Ñ Z Ñ Z Ñ 0 Since the map A Ñ X is the inclusion, the induced map has a left inverse, and the sequence k´1 splits, so H1pX,Aq – Z . If X “ S1 ˆ S1, then we have the exact sequence 2 B k 0 Ñ Z Ñ H1pX,Aq ÝÑ Z Ñ Z Ñ 0 22 2 k´1 By exactness, we have that ker B– Z and im B– Z . Since H1pX,Aq is a ﬁnitely generated abelian group with no torsion, it is determined by rank alone. Since its rank has to be k ` 1, k`1 we have that H1pX,Aq – Z . In both cases, we have that H2pX,Aq – Z. pbq X{A and X{B are both path-connected, so H0pX,Aq – H0pX,Bq – H0pX{Aq – 0. 2 2 2 2 4 H1pT _ T q – Z ‘ Z – Z by Corollary 2.25 in Hatcher. X{A is homotopy equivalent 2 2 2 2 4 to the wedge product T _ T , so H1pX,Aq – H1pX{Aq – H1pT _ T q – rZ . Similarly, 2 H2pX,Aq – Z . X{B is homotopy equivalent to the wedge product of a torus and a circle, 2 1 2 3 so H1pX,Bq – H1pX{Bq – H1pT _ S q – Z ‘ rZ – Z . Similarly,r H2pX,Bq – Z.

Exercise 65.rFor the subspacer Q Ă R, the relative homology group H1pR, Qq is free abelian and ﬁnd a basis.

Proof. Consider the exact sequence

B H1pRq Ñ H1pR, Qq ÝÑ H0pQq Ñ H0pRq

R is contractible, so H1pRq vanishes and H0pRq – Z. This gives B 0 Ñ H1pR, Qq ÝÑ H0pQq Ñ Z

By exactness, H1pR, Qq – ker B. H0pQq is an inﬁnite product of the integers, since Q has a path-component for each point in Q. Hence

B : Zq Ñ Z qPQ à where each Zq is a copy of Z. But the kernel of this map, for some basepoint q0 P Q, has basis eq ´ eq0 . Hence, since a subgroup of a free abelian group is free abelian, we have that H1pR, Qq is free abelian.

Exercise 66. HnpXq – Hn`1pSXq for all n, where SX is the suspension of X. More generally, thinking of SX as the union of two cones CX with their bases identified, compute the reducedr homologyr groups of the union of any finite number of cones CX with their bases identified.

Proof. Let p and q denote the tips of the two cones that compose SX. Let U “ SX{p and V “ SX{q. By Mayer-Vietoris, we have

... Ñ Hn`1pUq ‘ Hn`1pV q Ñ Hn`1pSXq Ñ HnpU X V q Ñ HnpUq ‘ HnpV q Ñ ... U X V “ X ˆ p0, 1q, which deformation retracts onto X. Additionally, both U and V deformationr retract ontor a point, thusr are contractible.r Hence ther Mayer-Vietorisr sequence gives the exact sequence

0 Ñ Hn`1pSXq Ñ HnpXq Ñ 0 Hence the two groups are isomorphic for all n. The second part of the propositionr follows by induction.r The induction hypothesis is

k k´1 k´1 Hn`1 CX – Hn`1 pSXq – HnpXq ˜i“1 ¸ i“1 i“1 ď à 23 à r r r The base case is given by the previous part, which also gives the second isomorphism of the inductive hypothesis. By the unnumbered example directly above Example 2.23 in Hatcher, CX{X – SX. The ﬁrst isomorphism in the inductive hypothesis then follows since

k k´1 k´1 k´1 Hn`1 CX – Hn`1 CX,X – Hn`1 SX – HnpXq ˜i“1 ¸ ˜i“1 ¸ ˜ i“1 ¸ i“1 ď ď ł à r r r r Exercise 67. Let f : pX,Aq Ñ pY,Bq be a map such that both f : X Ñ Y and the restriction f : A Ñ B are homotopy equivalences. (a) f˚ : HnpX,Aq Ñ HnpY,Bq is an isomorphism for all n. (b) For the case of the inclusion f : pDn,Sn´1q ãÑ pDn,Dn ´t0uq, f is not a homotopy equivalence of pairs. That is, there is no g : pDn,Dn ´ t0uq Ñ pDn,Sn´1q such that fg and gf are homotopic to the identity through maps of pairs. [ Observe that a homotopy equivalence of pairs pX,Aq Ñ pY,Bq is also a homotopy equivalence for the pairs obtained by replacing A and B by their closures .] Proof. paq The exact sequence of pairs coupled with the homotopy equivalences gives that

... HnpAq HnpXq HnpX,Aq Hn´1pAq Hn´1pXq ...

– – – –

... HnpBq HnpY q HnpY,Bq Hn´1pBq Hn´1pY q ...

But then the proposition for n ą 0 follows by the ﬁve-lemma. For n “ 0, this follows directly by the homotopy equivalences, as homotopy equivalent spaces have the same number of path components. pbq The observation in the problem statement follows since fpAq Ą fpAq. Suppose that the inclusion is a homotopy equivalence. But then by the observation in the problem statement and the previous part, this gives that there is a homotopy equivalence between Sn´1 “ Sn´1 and Dn ´ t0u “ Dn, a contradiction. Lemma 3. Chain homotopy of chain maps is an equivalence relation.

Proof. Let fn : Cn Ñ Dn be a chain map. Let ψn : Cn Ñ Dn`1 be given by ψnpσq “ 0. Then D C Bn ˝ ψn ` ψn´1 ˝ Bn´1 “ 0 “ fn ´ fn. Therefore, chain homotopy is reﬂexive. Let fn, gn : Cn Ñ Dn be two chain maps and let ψn : Cn Ñ Dn`1 be a chain homotopy D C f Ñ g such that fn ´ gn “Bn ˝ ψn ` ψn´1 ˝ Bn´1. Then ´ψ is a chain homotopy g Ñ f since D C D C gn ´ fn “ ´Bn ˝ ψn ´ ψn´1 ˝ Bn´1 “Bn ˝ p´ψnq ` p´ψn´1q ˝ Bn´1. Therefore, chain homotopy is symmetric. Let fn, gn, hn : Cn Ñ Dn be three chain maps and let ψn : Cn Ñ Dn`1 be a chain homotopy f Ñ g such that D C fn ´ gn “Bn ˝ ψn ` ψn´1 ˝ Bn´1 and let ϕn : Cn Ñ Dn`1 be a chain homotopy g Ñ h such that D C gn ´ hn “Bn ˝ ϕn ` ϕn´1 ˝ Bn´1. 24 Then adding the two previous chain homotopies gives a chain homotopy f Ñ h since D C D C fn ´ hn “ fn ´ gn ` gn ´ hn “Bn ˝ ψn ` ψn´1 ˝ Bn´1 ` Bn ˝ ϕn ` ϕn´1 ˝ Bn´1 D C “Bn ˝ pψn ` ϕnq ` pψn´1 ` ϕn´1q ˝ Bn´1 Therefore chain homotopy is transitive in addition to being reﬂexive and symmetric, and is thus an equivalence relation. Lemma 4. Chain homotopy is compatible with the composition of chain maps.

Proof. ψ be a chain homotopy f1 Ñ f2 and let ϕ be a chain homotopy g1 Ñ g2. Suppressing indices and composition for brevity, we have

g1f1 ´ g2f2 “ g1f1 ` g1f2 ´ g1f2 ´ g2f2

“ g1pf1 ´ f2q ` pg1 ´ g2qf2

“ g1pBψ ` ψBq ` pBϕ ` ϕBqf2

“Bg1ψ ` g1ψB ` Bϕf2 ` ϕf2B

“ Bpg1ψ ` ϕf2q ` pg1ψ ` ϕf2qB

Therefore g1f1 and g2f2 are chain homotopic.

Exercise 68. Let K(Ab) be Ch(Ab) with homotopy classes of maps as its morphisms. More precisely, let obpK(Ab)q be chain complexes in the category of abelian groups. Let hompK(Ab)q be chain maps up to homotopy. That is, two chain maps f, g : C Ñ D are equivalent if and only if there exists a sequence of morphisms ψn : Cn Ñ Dn`1 such D C that fn ´ gn “B ˝ ψn ` ψn´1 ˝ B . K(Ab) is a well-deﬁned category.

Proof. This follows by the previous two lemma. Exercise 69. A complex of abelian groups is called acyclic if its homology groups all vanish. (Similarly, a topological space is called acyclic if its associated singular chain complex is acyclic; i.e., if its singular homology groups all vanish.) A complex of abelian groups A is called contractible if the identity map on A is chain homotopic to the zero map. Prove that all contractible complexes are acyclic, and give an example of an acyclic complex that is not contractible.

Proof. Let A be a contractible chain complex. Then there exist maps ψn : An Ñ An`1 such that

σ “Bn`1ψn`1pσq ` ψnBnpσq.

We claim A is acyclic. Since im Bn`1 Ă ker Bn, it suﬃces to show ker Bn Ă im Bn`1. If σ P ker Bn, then σ “Bn`1ψn`1pσq ` ψnBnpσq “ Bn`1ψn`1pσq ` ψnp0q “ Bn`1ψn`1pσq, so σ P im Bn`1. The chain complex ˆ2 ... Ñ 0 Ñ Z ÝÑ Z Ñ Z{2Z Ñ 0 is acyclic but not contractible, since if it were contractible, the short exact sequence would split, and this is impossible since Z is isomorphic to the direct sum of Z and Z{2Z.

25 Exercise 70. S1 ˆ S1 and S1 _ S1 _ S2 have isomorphic homology groups in all dimensions, but their universal covers do not.

Proof. Since both spaces are path-connected, their homology groups match in dimension zero. 1 1 2 1 1 2 1 1 2 2 H1pS ˆ S q – Z and H1pS _ S _ S q – H1pS q ‘ H1pS q ‘ H1pS q – Z ‘ Z ‘ 0 – Z , so 1 1 1 1 2 their homology groups match in dimension one. H2pS ˆ S q – Z and H2pS _ S _ S q – 1 1 2 H2pS q ‘ H2pS q ‘ H2pS q – 0 ‘ 0 ‘ Z – Z, so their homology groups match in dimension two. In dimensions greater than two, the torus has trivial homology groups, and since S1 and S2 also have trivial homology groups in such dimensions and the homology group of the wedge product splits up as a direct sum, the homology groups of both spaces match in all dimensions. The universal cover of the torus is the plane, which is contractible. Therefore, it suffices to show that the universal cover of X of S1 _ S1 _ S2 has a nontrivial homology group of dimension greater than zero. Since X is simply-connected, H2pXq – π2pXq by the Hurewicz 1 1 2 Theorem. Since π2pXq – π2pS _ S r_ S q fl 0, we are done. r r r Exercise 71. Givenr a map f : S2n Ñ S2n, there exists x P S2n such that fpxq “ ˘x. 2n 2n Therefore, every map RP Ñ RP has a fixed point. The map h : S2n´1 Ñ S2n´1 given by hpx1, x2, . . . , x2nq “ px2, ´x1, x4, ´x3, . . . , x2n, ´x2n´1q 2n´1 2n´1 induces a map on RP Ñ RP with no fixed points. Proof. Suppose there does not exist x such that fpxq “ ´x. Then define a tangent vector field v on S2n as follows. For each point x P S2n, let ppfpxqq denote the stereographic projection of fpxq using ´x as the projection point. This is well defined since fpxq ‰ ´x for all x P S2n. Let vpxq “ ppfpxqq ´ x.

v is a continuous vector field since f and the stereographic projection are continuous, and since 2n is even, there must exist a point x0 such that v vanishes by Theorem 2.28 in Hatcher. This implies ppfpx0qq “ x0. But ppfpx0qq “ ppx0q, so by injectivity, fpx0q “ x0. 2n 2n 2n Let f : RP Ñ RP be any map. Extend f to a map g : S2n Ñ RP by gpxq “ 2n gp´xq “ fpqpxqq, where q is the natural projection S2n Ñ RP . By Proposition 1.33 in Hatcher, there exists a lift g : S2n Ñ S2n. By the argument above, there exists x such that gpx0q “ ˘x0. But then qpgpx0qq “ qpx0q, and since q ˝ g “ g, we have that gpx0q “ qpx0q. Since gpx0q “ fpqpx0qq, qpx0q is a fixed point of f. r 2n 2n r Let h be defined as in ther proposition. h descends to ar map RP Ñ RP since qphpxqq “ qphp´xqq. But since hpxq ‰ ˘x for any x, the descent map has no fixed points.

26 Exercise 72. paq If f : Sn Ñ Sn is a map of degree zero, then there exist x, y P Sn such that fpxq “ x, fpyq “ ý. pbq If F is a continuous, non-vanishing vector field on Dn, then there exist x PBDn where F points radially outward and y PBDn where F points radially inward. Proof. paq By property g of degree in Hatcher, since deg f ‰ p´1qn`1, it must have a fixed point. Following the proof of property g of degree, if fpxq ‰ ´x for any x, then the line segment fpxq Ñ x does not pass through the origin. Then define the homotopy p1 ´ tqfpxq ` tx f pxq “ t }p1 ´ tqfpxq ` tx} from f to the identity map. Since the identity map does not have degree zero, this gives the desired contradiction. n n´1 n´1 n´1 pbq Define G : D Ñ S by Gpxq “ F pxq{}F pxq}. The restriction G|BDn : S Ñ S n is equivalent to G ˝ i where i is the natural inclusion, so deg G|BDn “ 0 as D is contractible. The claim follows by the previous argument. Exercise 73. paq The long exact sequence of homology groups associated to the short n exact sequence of chain complexes 0 Ñ CipXq ÝÑ CipXq Ñ CipX; Z{nZq yields short exact sequences

0 Ñ HipXq{nHipXq Ñ HipX; Z{nZq Ñ n-TorsionpHi´1pXqq Ñ 0 where n-TorsionpGq is the kernel of the map G ÝÑn G, g ÞÑ ng. pbq HipX; Z{pZq “ 0 for all i and all primes p if and only if HipXq is a vector space over Q for all i. r r Proof. paq We have the long exact sequence

η fn ... Ñ HipXq ÝÑ HipXq ÝÑ HipX; Z{nZq Ñ ...

Since im η “ ker f, we have that ker f “ nHipXq. By the first isomorphism theorem, HipXq{ ker f – HipX; Z{nZq. Additionally, im B“ ker fn´1 “ n-TorsionpGq, we have the sequence 0 Ñ HipXq{nHipXq Ñ HipX; Z{nZq Ñ n-TorsionpHi´1pXqq Ñ 0 which is exact since the first map is injective and since Bf “ 0. pbq ( ùñ ) It suffices to show that HipXq is free, abelian, torsion-free, and infinitely- generated. ( ðù ) Q has no torsion and HipX; Qrq{nHipX; Qq “ 0, so by the splitting lemma, the claim follows. 8 Exercise 74. HipRP ; Z{2Zq – Z{2Z for all i P N .

8 Proof. There exists a two-sheeted cover p : S8 Ñ RP , so by the proof of proposition 2B.6 in Hatcher, the (transfer) sequence 8 8 8 8 ... Ñ HipRP ; Z{2Zq Ñ HipS , Z{2Zq Ñ HipRP ; Z{2Zq Ñ Hi´1pRP ; Z{2Zq Ñ ... is exact. Since S8 is contractible, the sequence is 8 8 8 ... Ñ HipRP ; Z{2Zq Ñ 0 Ñ HipRP ; Z{2Zq Ñ Hi´1pRP ; Z{2Zq Ñ 0 Ñ ... 27 In particular, we have short exact sequences 8 8 0 Ñ Hi`1pRP ; Z{2Zq Ñ HipRP ; Z{2Zq Ñ 0 8 8 for all i ą 0. Therefore Hi`1pRP ; Z{2Zq – HipRP ; Z{2Zq for all i ą 0. 8 8 We finish the remaining two cases. RP is path-connected, so H0pRP ; Z{2Zq – Z{2Z. Adding cells of dimension greater than two does not affect the first homology group, so 8 2 H1pRP ; Z{2Zq – H1pRP ; Z{2Zq – Z{2Z.

Exercise 75. Let f : C Ñ D be a morphism of chain complexes. Deﬁne Cf as the mapping cone of f such that pCf qn “ Cn´1 ‘ Dn, with boundary maps dpx, yq “ C D p´dn´1pxq, fn´1pxq ` dn pyqq. Cf is a chain complex.

Cf Cf Proof. We check that dn ˝ dn`1 “ 0. We have C Cf ´dn´1 0 dn “ D fn 1 d ˆ ´ n ˙ C C Cf Cf dn´1 ˝ dn 0 dn ˝ dn`1 “ C D D D ´fn 1 ˝ d ` d ˝ fn d ˝ d ˆ ´ n n n n`1˙ The diagonal terms are zero by assumption. The last entry is zero by the commutativity of

C dn ... Cn Cn´1 ...

fn fn´1 . D dn ... Dn Dn´1 ...

Exercise 76. Deﬁne a map j : D Ñ Cf by jnpyq “ p0, yq, and a map d : Cf Ñ Cr´1s n by dnpx, yq “ p´1q x. Here Cr´1s denotes the same complex as C, but re-indexed so Cr´1s C that Cr´1sn “ Cn´1 and B “B . j d 0 Ñ D ÝÑ Cf ÝÑ Cr´1s Ñ 0 is an exact sequence of complexes.

Proof. If y P ker jn, then y “ 0. Hence, ker j “ 0. If px, yq P ker d, then x “ 0. Hence, im j “ ker d. d is surjective, and thus the sequence is exact. Exercise 77. A map f : C Ñ D of complexes is a quasi-isomorphism if it induces isomorphisms on all homology groups. A morphism f is a quasi-isomorphism if and only if Cf is acyclic.

Proof. p ùñ q Let rpx, yqs P HnpCf q. If px, yq is a cycle, then dpxq “ 0 and dpyq ` fpxq “ 0. fpxq “ dpýq, so fpxq is a boundary, and by injectivity, x is also a boundary, so x “ dpx1q and rxs “ 0. This gives dpy `fpx1qq “ 0, so y `fpx1q is a cycle. By surjectivity, ry `fpx1qs “ fprx2sq, so y “ fpx2 ´ x1q. Thus dpx, yq “ p´dpx2 ´ x1q, fpx2 ´ x1qq “ px2 ´ x1, 0q so px, yq is a boundary. Since every cycle is a boundary, Cf is acyclic as all of its homology groups vanish. 28 p ðù q Suppose f˚rxs “ 0. Since r´x, ys is a cycle, it is also a boundary as Cf is acyclic. In particular x is a boundary, and hence f˚ is injective since its kernel is trivial. Now suppose that rys is a cycle in Dn. Then p0, yq is a cycle, and hence a boundary as Cf is acyclic. Hence 1 1 rys “ rdpy q ` `fpx qs and hence in the image of f˚, so f is a quasi-isomorphism. Exercise 78. The homology groups of the following 2-complexes are described in the following proof. paq The quotient of S2 obtained by identifying north and south poles to a point. pbq S1 ˆ pS1 _ S1q. pcq The space obtained from D2 by first deleting the interiors of two disjoint sub-disks in the interior of D2 and then identifying all three resulting boundary circles together via homeomorphisms preserving clockwise orientations of these circles. 1 1 1 pdq The quotient space of S ˆ S obtained by identifying points in the circle S ˆ tx0u 1 that differ by a 2π{m - rotation and identifying points in the circle tx0u ˆ S that differ by a 2π{n - rotation.

Proof. paq By Example 0.8 in Hatcher, X » S2 _ S1. By Corollary 2.25 in Hatcher,

2 1 Z n “ 0, 1, 2 HnpXq – HnpS _ S q “ . #0 n ą 2 pbq Let S1 and S1 _ S1 have the CW structures

Figure 2. CW complex structure on S1

Figure 3. CW complex structure on S1 _ S1 29 By Theorem A.6 in Hatcher, the cellular chain complex of the product is

d“0 d“0 0 Ñ Ztγ ˆ δ1, γ ˆ δ2u ÝÝÑ Ztα ˆ δ1, α ˆ δ2, β ˆ γu ÝÝÑ Ztα ˆ βu Ñ 0 Therefore, Z n “ 0 3 $Z n “ 1 HnpXq – 2 ’Z n “ 2 &’ 0 n ą 2 c Let X have the following CW complex structure.’ p q %’

Figure 4. CW complex structure on X

Then the cellular chain complex of X is

d2 0 0 Ñ ZtV u ÝÑ Ztγ, δ1, δ2u ÝÑ Ztαu Ñ 0

where d2pV q “ ´γ. Hence, Z n “ 0 2 HnpXq “ $Z n “ 1 &’0 n ą 1 pdq We modify the attachment of the two-cell of the usual CW complex structure of the torus %’ to maintain the identiﬁcation. Given the zero-cell α, and the two one-cells δ1, δ2, attach the n m ´n ´m two-cell along δ1 δ2 δ1 δ2 . Thus the cellular chain complex for X is 0 0 0 Ñ ZtV u ÝÑ Ztδ1, δ2u ÝÑ Ztαu Ñ 0 30 Hence, Z i “ 0 Z2 i “ 1 HipXq “ $ ’Z i “ 2 &’ 0 i ą 2 ’ %’ n Exercise 79. If X is a CW complex, then HnpX q is free.

n n´1 n´1 n´2 n n´1 Proof. Consider the cellular map dn : HnpX ,X q Ñ Hn´1pX ,X q. Each HnpX ,X q n is free with generators the n-cells of X, so it suﬃces to show that ker dn – HnpX q, as the kernel of a homomorphism is a subgroup, and a subgroup of a free group is free. By exactness of the cellular chain complex, im dn`1 “ ker dn. Additionally, by Hatcher p. 139, the diagram

n HnpX q

Bn`1 jn

n`1 n dn`1 n n´1 Hn`1pX ,X q HnpX ,X q

n n n´1 commutes. Hence, HnpX q is a subgroup of HnpX ,X q, so is free.

Exercise 80. Suppose the space X is the union of open sets A1,...,An such that each

inter-section Ai1 X...XAik is either empty or has trivial reduced homology groups. Then HipXq “ 0 for i ě n ´ 1, and give an example showing this inequality is best possible, for each n.

r k k Proof. Let Xk “ i“1 Ai and Yk “ i“1 Ai. We claim for all k in t1, 2, . . . , nu, HipXk XYk`1q vanishes for all i ą k ´ 2. The proposition follows from the case k “ n. Ť Ş The case k “ 1 follows by assumption. By induction, we have that HipXk´1rX Yk`1q “ 0 for all i ą k ´ 3. This gives the sequence r HipXk´1 X Ykq Ñ HipXk´1 X Yk`1q ‘ HipYkq Ñ HipXk´1 X Yk`1q Ñ Hi´1pXk´1 X Ykq by Mayer-Vietoris. This yields the exact sequence r r r r r HipXk´1 X Yk`1q Ñ HipXk X Yk`1q Ñ HipXk´1 X Ykq Since the outer terms are zero by induction for all i ą k ´ 2, the proposition follows. Consider any n ąr 3 and let X “ Srn´2. Decompose rSn´2 into n open sets such that the conditions of the proposition hold. Since Hn´2pXq “ Z, the proposition gives the best possible bound. r Exercise 81. Let F be a free abelian group, and consider the solid diagram F

h f

A g B Then there exists a map h making the diagram commute. 31 Proof. Let tgiu be a basis of F . Since g is surjective, for all fpgiq, there exists ai such that gpaiq “ fpgiq. Let hpgiq “ ai. By the universal property of free groups, h extends uniquely to a homomorphism F Ñ A. But then since g ˝h and f agree on where they map generators, they must agree on all of F , again by the universal property and uniqueness. Exercise 82. If f : A Ñ F is a surjective map of abelian groups with F free, then A “ ker f ‘ F 1, where F 1 – F . Proof. The short sequence f 0 Ñ ker f ÝÑi A ÝÑ F Ñ 0 is exact since the inclusion i is injective and f is surjective. Now consider the solid diagram F g id A F f By the previous proposition, there exists a homomorphism g such that f ˝ g “ id. But then the short exact sequence splits, and A – ker ‘F , so the proposition follows.

Exercise 83. Let pC, dq be a chain complex of abelian groups such that each Cn is free. Then C is quasi-isomorphic to the chain complex H with Hn “ HnpCq and all diﬀerentials the zero map. Equivalently, C is formal or quasi-isomorphic to its homology. Proof. Consider the solid diagram

dn`1 dn ... Cn`1 Cn Cn´1 ...

φn`1 φn φn´1

... ker dn`1 0 ker dn 0 ker dn´1 ... im dn`2 im dn`1 im dn We claim there exists a chain map φ : C Ñ H such that the diagram commutes and φ induces isomorphisms on all homology groups. Explicitly, φn ˝dn`1 “ 0 and Hnpφq : HnpCq Ñ HnpHq is an isomorphism for all n. Consider the short exact sequence

0 Ñ ker dn Ñ Cn Ñ im dn Ñ 0.

By the previous proposition, it splits, and hence Cn – ker dn ‘ im dn by some isomorphism ψn. Let π1 : ker dn ‘ im dn Ñ ker dn be the natural projection onto the ﬁrst component, and let qn : ker dn Ñ ker dn{ im dn`1 be the quotient map to the coset space. Finally, let φn “ qn ˝ π1 ˝ ψn. ker dn φn ˝ dn 1 “ 0 as im dn 1 Ă ker qn, so φ is a valid chain map. Note that HnpHq “ ` ` im dn`1 since each diﬀerential is the zero map. Hence, the induced map

ker dn ker dn Hpφnq : Ñ im dn`1 im dn`1

is an isomorphism since ψn is an isomorphism. 32 Exercise 84. A map f : X Ñ Y between connected, n-dimensional CW complexes is a homotopy equivalence if it induces an isomorphism on πi for all i ď n.

Proof. Let py : Y Ñ Y denote the covering map of the universal cover of Y , and let px : X Ñ X denote the covering map of the universal cover of X. Now consider the map f˝px : X Ñ Y . Yr is a connected CW complex, which is locally contractible, and in particular, path-connectedr and locally path-connected. Since X is simply connected, pf ˝pxq˚pπ1pXqq – 0 Ă ppyqr˚pπ1pY qq, and thus by the lifting criterion, the diagram r r f p X ˝ x Y Ć px py r r X Y f commutes. px and py induce isomorphisms on πi for i ą 1, and f also induces isomorphisms on πi for all i ď n by assumption. Therefore, by commutativity of the diagram, f ˝ px induces isomorphisms on πi for 1 ă i ď n. Additionally, since the universal covers are simply connected, f ˝ px induces isomorphisms on πi for i ď n. Č Since Y is a deformation retract of M , we may replace Y with the mapping cone and f˝px Č regard f ˝ px as an inclusion. Since X and Y are simply connected, π1pY, Xq “ 0. By r Ć r the relative Hurewicz theorem, the ﬁrst nonzero πipY, Xq is isomorphic to the ﬁrst nonzero

HipY, XČq. By the long exact sequencer of homotopy,r since f ˝ px induces isomorphismsr r up to πn, the groups πipY, Xq vanish up to i “ n, andr sor also then do the groups HipY, Xq. Hencer therer are induced isomorphisms on all homology groupsČHi for i ď n. Now the claim thatrXrand Y are n-dimensional CW complexes suffices to finish offr ther proof of the proposition. This is because if the claim is true, then HipY, Xq vanish for

i ą n, and so by Corollaryr 4.33r in Hatcher, f ˝ px is a homotopy equivalence. But then it must induce isomorphisms on all homotopy groups πi for all i ą n, and hencer r so must f by commutativity of the diagram. The propositionČ then follows by Whitehead’s theorem. The claim itself regarding universal covers of CW complexes is proven in detail in Whitehead, J. H. C. Combinatorial homotopy. I. Bull. Amer. Math. Soc. 55 (1949), 213–245. Each attaching map of the base CW complex can be lifted to the universal cover as cells are contractible, and so there is one cell per each lift corresponding to the base space. This cell decompositon has the right topology since the covering map is a local homeomorphism.

Exercise 85. Let X be an pn ´ 1q-connected CW complex with n ą 1. n`1 n`1 paq The natural maps πn`1pX q Ñ πn`1pXq and Hn`1pX q Ñ Hn`1pXq are surjective. n`1 n`1 pbq The Hurewicz map πn`1pX q Ñ Hn`1pX q is surjective. pcq The map πn`1pXq Ñ Hn`1pXq is surjective. pdq If n “ 1, the corresponding statement is false.

Proof. paq Consider an element γ P πn`1pXq. By cellular approximation, it can be represented by a cellular map Sn`1 Ñ X. Hence, we can ﬁnd a representative which factors over 33 the inclusion Xn`1 Ñ X, thus γ lies in the image of the induced map of the inclusion. The case for homology follows by Lemma 2.34 in Hatcher. pbq Consider the commutative diagram

n a n`1 b n`1 n c n πn`1pX q πn`1pX q πn`1pX ,X q πnpX q

h1 h2 h3 h4

n a1 n`1 b1 n`1 n c1 n Hn`1pX q Hn`1pX q Hn`1pX ,X q HnpX q given by the long exact sequence for the pair pXn`1,Xnq associated to the Hurewicz map n 1 h. By cellular homology, Hn`1pX q vanishes, so a is injective. Furthermore, h3 and h4 are isomorphisms by the Hurewicz theorem. The claim follows by a diagram chase. Consider n`1 ´1 1 1 1 1 any element y in πn`1pX q. Let k “ h3 pb pyqq. Now h4pcpkqq “ c ph3pkqq “ c pb pyqq “ 0 by commutativity and exactness. By injectivity of h4, cpkq is trivial, so by exactness, there 1 1 exists x such that bpxq “ k. By commutativity, h3pbpxqq “ b ph2pxqq “ b pyq. By injectivity 1 of a , h2pxq “ y. n`1 n`1 n`1 pcq The composition of πn`1pX q Ñ Hn`1pXq and πn`1pX q Ñ Hn`1pX q is a surjective map, which is a restriction of h, so h must be surjective. pdq The torus, which is 0-connected, has contractible universal cover, and hence trivial homotopy. But H2pTq – Z as shown previously, so no surjection exists.

Exercise 86. The tensor product of two chain complexes C b C1 is a chain complex, where the diﬀerential map is given by 1 1 BCbC pc, c1q “ pBC pcq, c1q ` p´1qdegpcqpc, BC pc1qq.

Proof. Composing twice, we have

1 B2pc b c1q “ BpBC pcq b c1 ` p´1qic b BC pc1qq 1 1 “ p´1qi´1BC pcq b BC pc1q ` p´1qiBC pcq b BC pc1q “ 0

Exercise 87. Let F be a field and X be a space such that HipX; F q has finite dimension for all i. Define the Poincaréseries pX to be the formal power series i pX ptq “ pdimF HipX; F qqt . i ÿ n n n 8 8 The proof below gives formulas for the Poincaréseries of S , RP , CP , RP , CP , and the orientable surface Mg of genus g. If X and Y are spaces with well-defined Poincaré series, then pX Y ptq “ pX ptq ` pY ptq and pX Y ptq “ pX ptq ` pY ptq ´ 1.

š Ž

n n F i P t0, nu n Proof. S : HipS ; F q “ . Hence, pSn ptq “ 1 ` t . #0 else 34 n n 1 ` t n odd n RP : We have pRP ptq “ . This follows since for n even, all homology #1 else groups of dimension greater than zero with coefficients in a 2-divisible ring vanish, while for n odd, the n-th homology group does not vanish as we have shown previously. n CP : Complex projective space has homology 0 in all odd dimensions and homology n n n 2n HipCP ; Z “ Z in all even dimension up to 2n. Hence, pCP ptq “ i“0 t . 8 RP : All homology groups are with coefficients in Z{2Z are Z{2Z, and hence pRP8 ptq “ 8 n ř i“0 t . 8 8 2n 8 CP : pRP ptq “ i“0 t by Hatcher’s description of the homology being Z{2Z in each evenř dimension. ř Mg: By using the identification with the connected sum of tori, we know the homology 2g groups are Z for H0pM; Zq and H2pM; Zq. Additionally, H1pM; Zq “ Z . Hence, pMg ptq “ 1 ` 2gt ` t2. For pX Y ptq “ pX ptq ` pY ptq, we know that HipX Y ; F q “ HipX; F q ‘ HipY ; F q, and thus the dimensions add and the result follows by linearity. š š Additionally, HipX _ Y ; F q “ HipX Y ; F q. This gives the same result for i ą 0, but for i “ 0, it vanishes, so pX Y ptq “ pX ptq ` pY ptq ´ 1. š Ž r Exercise 88. If F is a field, the Künnethformula reduces to a natural isomorphism

HnpX ˆ Y ; F q – HppX; F q bF HqpY ; F q. p`q“n à If X and Y are spaces with well-deﬁned Poincar´eseries, then pXˆY ptq “ pX ptqpY ptq.

Proof. The formula reduces since if F is a ﬁeld, the groups TorF vanish, giving that the map in the short exact sequence in the proposition is an isomorphism. The Poincar´eformula follows since

dimF pHnpX ˆ Y ; F qq “ dimF pHppX; F q b HqpY ; F qq p`q“n ÿ “ dimF HppX; F q ¨ dimF HqpY ; F q p`q“n ÿ by the fact that the dimension of a product is the product of the dimensions. But this is exactly the formula for the coeﬃcients of the product of two polynomials, hence pXˆY ptq “ pX ptqpY ptq.

Exercise 89. Let X be the CW complex obtained by attaching two 2-cells to S1, one via a degree p map and one via a degree q map. Below is a description of the homology of X and when X is equivalent to S2.

Proof. Taking care of the homology groups independent of p and q, note that HnpXq “ 0 for n ą 2 since X has no n-cells for n ą 2. Additionally, since X is non-empty and path- connected, H0pXq » Z. Now for some notational setup. Give S1 the standard CW complex structure of one 0-cell 0 1 2 2 e and one 1-cell e with the usual attaching map, and let ep and eq denote the two 2-cells attached via a degree p map and a degree q map, respectively. X is connected and has only one 0-cell, so the diﬀerential map d1 “ 0 by Hatcher, p. 140. Furthermore, by assumption, 35 2 1 2 1 d2pepq “ pe and d2peqq “ qe . The cellular chain complex of X is

2 2 d2 1 d1 0 0 Ñ Z ‘ Ztep, equ ÝÑ Zte u ÝÑ Zte u Ñ 0.

Let’s pin-down the kernel and image of d2. The trivial case is if the degrees of both 2 2 attaching maps are zero, that is, p “ q “ 0. Then d2pmep ` neqq “ 0 ` 0 “ 0, so H1pXq “ 2 2 kerpd1q{ impd2q » Z{0 » Z and H2pXq “ kerpd2q » Z . In this case, X and S cannot be (weakly) homotopy equivalent since their homology groups are not isomorphic. 2 2 2 Otherwise, suppose either p ‰ 0 or q ‰ 0. Any pair pm, nq P Z maps as d2pmep ` neqq “ 2 2 1 1 1 md2pepq ` nd2peqq “ mpe ` nqe “ pmp ` nqqe , using the homomorphism property. By Bézout’sIdentity, all integers of the form pmp ` nqq are multiples of pp, qq, the greatest common divisor of p and q. Hence, im d2 » pp, qqZ, and H1pXq » Z{pp, qqZ. If pp, qq ‰ 1, 2 2 then X is not homotopy equivalent to S since H1pXq » Z{pp, qqZ fi 0 “ H1pS q. Now to understand the kernel. By the above, the kernel of d2 is given by all points pm, nq P Z‘Z such that pm “ ´qn, which is generated by pq{pp, qq, ´p{pp, qqq since pq{pp, qq is the least common multiple of p and q. Hence, ker d2 » Z and thus H2pXq “ ker d2 » Z. If pp, qq “ 1, we claim that we can construct a map f : X Ñ S2 which induces an isomorphism on the homology groups, and thus is a homotopy equivalence. It is necessary to note a small, technical detail here. We have only shown ‘Whitehead’s Theorem for homology’, Corollary 4.33 in Hatcher, for singular homology. Therefore, we need f to induce an isomorphism on singular homology. However, if f induces an isomorphism on cellular homology, and f is a cellular map, then the induced map corresponds naturally to the isomorphism between cellular and singular homology, so a cellular map suffices. This is exercise 2.2.17 in Hatcher, which follows by the naturality of the long exact sequence of homology, the proof of which is not included here for brevity. We proceed by constructing the claimed cellular map. For one, the induced isomorphism on H2 must be an isomorphism between the kernels of the differential maps, since no cells of dimension three or greater are attached. The kernel of d2 is generated by pq, ´pq in this case since p and q are relatively prime by the argument 1 2 above, and the kernel of d2 is Z since this differential is trivial for S using the CW complex structure of one 2-cell and one 0-cell. Therefore, the induced map must send pq, ´pq ÞÑ 1. Then, to create the cellular map which induces the desired map, let m, n be the Bézout coefficients, the two integers such that pm`qn “ 1 “ pp, qq. Let e2 be the 2-cell of S2. Then 2 2 2 2 the cellular map is given by mapping ep ÞÑ e via any degree n map, and mapping eq ÞÑ e via any degree ´m map. This gives that f˚pq, ´pq “ qn ´ p´mqp “ 1. f can also map the 0-cell of X to the 0-cell of S2, and the 1-cell of X to the 0-cell of S2 or any other point, thus inducing isomorphisms on H1 and H0 as well. f thus induces isomorphisms on the cellular homology groups, and since it is cellular, its induced map also induces an isomorphism on the singular homology groups, so it must be a homotopy equivalence by Corollary 4.33 in Hatcher.

Exercise 90. Any continuous f : Sn Ñ Sn such that deg f ‰ p´1qn`1 has a ﬁxed point.

Proof. This is the contrapositive of property g of degree in Hatcher, p. 134. Below is a reproduction of the proof for completeness. Suppose that f has no ﬁxed point. Then the line segment γ : I Ñ Rn`1 from fpxq to ´x given by γtpxq “ p1 ´ tqfpxq ´ tx does not pass through the origin. Thus, there exists a 36 homotopy

γtpxq ftpxq “ }γtpxq}

from fpxq to the antipodal map. But by property f of degree in Hatcher, p. 134, the antipodal map has degree p´1qn`1 since it is the composition of pn ` 1q reﬂections in Rn`1, each changing the sign of one coordinate. Hence, by property c of degree in Hatcher, deg f “ n 1 p´1q ` .

Exercise 91. Recall than an equivalence of categories is a functor F : C Ñ C1, a 1 functor G : C Ñ C, and two natural isomorphisms : FG Ñ idC1 and η : idC Ñ GF . A natural isomorphism is a natural equivalence such that each deﬁning morphism is an isomorphism, and idC and idC1 are the identity functors. Suppose that C has all limits. Then C1 has the same property and if α : D Ñ C is a diagram in C, indexed by a small category D, limpF αq “ F plim αq in the sense that for any choice of the limits, the results are uniquely isomorphic.

Proof. For all diagrams α1 : D1 Ñ C1 in C1, Gα1 : D1 Ñ C yields a diagram in C. By assumption, C has a limit pL, ψq of Gα1. We claim F ppL, ψqq is a limit of α1. First, we must show it is a cone. For all objects X and Y in D1, and all morphisms f : X Ñ Y , the diagram

L (1) ψX ψY pGα1qpXq pGα1qpY q pGα1qpfq

commutes by functoriality of Gα1 and the fact that pL, ψq is a limit of Gα1, and in particular, a cone. By functoriality of F , the diagram

F pLq (2) F pψX q F pψY q

pFGqpα1pXqq pFGqpα1pY qq pFGqpα1pfqq

commutes. Now, since is a natural isomorphism, it is in particular a natural transformation, so the diagram

pFGqpα1pfqq pFGqpα1pXqq pFGqpα1pY qq

(3) α1pXq α1pY q

1 1 1 1 idC pα pXqq 1 idC pα pY qq idC1 pα pfqq 37 α1pfq commutes. The bottom row of 3 can simply be replaced by α1pXq ÝÝÝÑ α1pY q by the deﬁnition of idC1 . Combining 3 and 2 together yields the commutative diagram

F pLq F pψX q F pψY q

pFGqpα1pfqq (4) pFGqpα1pXqq pFGqpα1pY qq .

α1pXq α1pY q α1pXq α1pY q α1pfq

Thus, F pLq is a cone to α1. Suppose pM, φq is another cone to α1. This yields the commutative diagram

φX φY

(5) F pLq .

F pψX q F pψY q

pFGqpα1pfqq pFGqpα1pXqq pFGqpα1pY qq

α1pXq α1pY q α1pXq α1pY q α1pfq

By functoriality of G applied to the outer morphisms of 5,

GpMq (6) GpφX q GpφY q .

Gpα1pXqq Gpα1pY qq Gpα1pfqq

commutes. Now, since pL, ψq is a limit of Gα1, there exists a unique map u such that

GpMq

GpφX q u GpφY q (7) L . ψX ψY

Gpα1pXqq Gpα1pY qq Gpα1pfqq 38 commutes. By functoriality of F , diagram 7, and diagram 5, the diagram

φ φ X pFGqpMq Y

pFGqpφX q F puq pFGqpφY q (8) . F pLq

F pψX q F pψY q

pFGqpα1pfqq pFGqpα1pXqq pFGqpα1pY qq

α1pXq α1pY q α1pXq α1pY q α1pfq commutes. Flattening out right square of 8 yields the solid commutative diagram

φ M Y α1pY q

(9) M α1pY q pFGqpφ q pFGqpMq Y pFGqpα1pY qq where the map M exists making the diagram commute since is a natural transformation ´1 FG Ñ idC1 . Additionally, M has an inverse M since is also a natural isomorphism. By the same argument, we have commuativity of the left square. Thus, comibining these facts, and diagrams 9 and 8 yields the commutative diagram

´1 M M

φ φ X pFGqpMq Y

pFGqpφX q F puq pFGqpφY q (10) . F pLq

F pψX q F pψY q

pFGqpα1pfqq pFGqpα1pXqq pFGqpα1pY qq

α1pXq α1pY q α1pXq α1pY q α1pfq 39 ´1 This yields that the mediating map of the limit, F puqM : M Ñ F pLq always exists. It remains to show that this map is unique. Any map µ : M Ñ F pLq factors through pFGqpMq using M , since

µ M F pLq

µ ´1 M M pFGqpMq

commutes. Thus, if the map pFGqpMq Ñ F pLq is unique, then µ is unique. Hence, it suﬃces to show that the map F puq is unique in 10. Suppose there exists µ such that

pFGqpMq

pFGqpφX q µ pFGqpφY q

(11) F pLq .

F pψX q F pψY q

pFGqpα1pfqq pFGqpα1pXqq pFGqpα1pY qq

commutes. We aim to show µ “ F puq. By functoriality of G, and application of the natural isomorphism η, and the fact that pL, ψq is a limit of Gα1, the diagram

GpMq

η ´1 M ηM pGF GqpMq

GpφX q GpφY q Gpµq

pGF GqpφX q pGF qpLq pGF GqpφY q (12) . ηL pGF qpψ q pGF qpψ q X L Y

pGF qpGpα1pfqqq pGF qpGpα1pXqqq pGF qpGpα1pY qqq

ηGpα1pXqq ηGpα1pY qq Gpα1pXqq Gpα1pY qq Gpα1pfqq

´1 commutes. Note that ηLGpµqηM : GpMq Ñ L is a mediating map for pL, ψq, so by uniqueness ´1 of the limit, we must have that u “ ηLGpµqηM . This equation is just the commutative 40 diagram

Gpµq pGF GqpMq pGF qpLq

(13) ηM ηL GpMq u L We also have the commutative diagram of the natural transformation

pGF qpuq pGF GqpMq pGF qpLq

(14) ηM ηL GpMq u L

By the previous two diagrams, ηLGpµq “ ηLGpF puqq so Gpµq “ GpF puqq as ηL is an isomorphism. Composing with F on both sides yields pFGqpµq “ pFGqpF puqq. Now, using the natural transformation gives the commutative diagrams

pFGqpµq pF GF GqpMq pF GF qpLq

pFGqpMq F pLq µ pFGqpMq F pLq

pF GF qpuq pF GF GqpMq pF GF qpLq

pFGqpMq F pLq F puq pFGqpMq F pLq

Writing commutativity down explicitly, F pLqpFGqpµq “ µpFGqpMq and F pLqpFGqpF puqqq “ F puqpFGqpMq. Since pFGqpµq “ pFGqpF puqq, we have that F puqpFGqpMq “ µpFGqpMq, or that µ “ F puq by composing on the left with pFGqpMq. Thus the mediating map is unique, and F ppL, ψqq is a limit of α1, and thus C1 has all limits. We now show that F plim αq “ limpF αq. This follows directly from two claims: paq If A and B are limits over the same diagram, then there is a unique isomorphism A Ñ B. pbq limpF αq and F plim αq are limits over the same diagram. We have already proven claim paq in class, using the universal property and uniqueness of the mediating map for limits. For claim pbq, limpF αq is a limit over F α by deﬁnition. We have shown above that given the equivalence of categories, if pL, ψq is a limit of α, then F ppL, ψqq is a limit of F α.

Exercise 92. Let X denote pS1q3, the three-dimensional torus, with its natural product CW complex structure. Let f : S3 Ñ S2 be the Hopf ﬁbration and g : X Ñ S3 the map collapsing the two-skeleton of X to a point. f ˝ g induces the trivial map on all πn and all Hn, but is not nullhomotopic.

1 1 3 1 1 1 Proof. Since S is path-connected, πnppS q q » πnpS q ˆ πnpS q ˆ πnpS q by Proposition r 1 3 1 4.2 in Hatcher. Thus, πnppS q q is trivial for n ‰ 1 since πnpS q “ 0 for n ‰ 1, and thus 41 1 3 2 the induced map pf ˝ gq˚ : πnppS q q Ñ πnpS q is trivial for all n ‰ 1. But for n “ 1, 2 3 π1pS q “ 0, so the induced map must be trivial for all n. Note that HnpS q “ 0 for n ‰ 3, 2 3 2 and HnpS q “ 0 for n ‰ 2, so the induced map f˚ : HnpS q Ñ HnpS q must be trivial for all n, and hence the induced map pf ˝ gq˚ must also be trivial for all n. r Supposer for a contradiction that f ˝g is null-homotopicr via ar homotopy h : pS1q3 ˆI Ñ S2 2 where hpx, 0q “ pf ˝ gqpxq and hpx, 1q “ s0 for some s0 P S . Since f is a ﬁbration, there exists a lift h such that g pS1q3 S3 r h pS1q3ˆt0u f r S1 3 I S2 p q ˆ h

commutes. By commutativity, f ˝ h “ h, and since hpx, 1q “ s0, it follows that hpx, 1q is ´1 1 1 3 1 3 contained in f ps0q “ S . Thus ph1q˚ factors as a map pS q Ñ S Ñ S , but the ﬁrst map 1 must induce the trivial map on ther homology groups of dimension three since H3prS q “ 0. 1 3 3 Hence, g˚ : H3ppS q q Ñ H3pS q mustr be trivial, since g “ h0 is homotopic to h1 via h. But g takes the 3-cell of pS1q3 to the 3-cell of S3 via a degree one map, and since it is cellular, the induced a map on the cellular chain complexes cannotr be trivial due tor its degreer as 1 3 3 H3ppS q q » Z by Hatcher, p. 143, and H3pS q » Z. Exercise 93. Let i : A Ñ X be an inclusion. Show that i is null-homotopic if, and only if, X is a retract of the mapping cone Ci of i. If i is nullhomotopic, HnpX,Aq »

HnpXq ‘ Hn´1pAq for each n ě 1. Proof. To codify conventions, let C be the space A I X with the identifications a, 0 r r i ˆ p q „ pa1, 0q for all a, a1 P A, and pa, 1q „ ipaq “ a. š ( ùñ ) Let h : A ˆ I Ñ X be the null-homotopy of the inclusion such that hpa, 0q “ a0 and hpa, 1q “ a for some a0 P A. To construct the necessary retract, we can first define a map r : A ˆ I X Ñ X which is constant on equivalence classes, and thus descends to a retract r : Ci Ñ X. Let r be such that if x P X, rpxq “ x. Otherwise, if x “ pa, tq P AÎ, rpxq “ rpa,š tq “ hpa, tq. To verify that the descent map is well-defined, we check the two r 1 1 identifications. rpa, 0q “ hpa,r0q “ a0 “ hpa , 0q “ rpa r, 0q and rpa, 1q “ hpa, 1q “ a “ rpaq, so ther descentr map is well-defined. It is also a retract since rpxq “ x for all x P X. ( ðù ) Let r r: Ci Ñ X be a retract. This extendsr to a maprr : A ˆ I X Ñ X whichr is constant on equivalence classes pa, 1q „ a and pa, 0q “ pa1, 0q for all a, a1 P A. Hence, for 1 1 š all a, a P A, rpa, 0q “ rpa , 0q, thus rpa, 0q “ a0 for some a0 P A forr all a P A. Additionally, rpa, 1q “ rpaq, and since r is a retract, rpaq “ a. Recapitulating, for all a P A, rpa, 0q “ a0 and rpa, 1q “ra, so r|ArÎ is the desiredr null-homotopy of the inclusion i : A Ñ X. r Now, supposer that i : A Ñ X is null-homotopic.r Consider the long exact sequencer of a pair r r i˚ j˚ B ... Ñ HnpAq ÝÑ HnpXq ÝÑ HnpX,Aq ÝÑ Hn´1pAq Ñ ....

Since i is null-homotopic, i˚ “ 0 for all HnpAq Ñ HnpXq for n ě 1, and the identity for n “ 0. By exactness, each j˚ has trivial kernel, and thus is injective for n ě 1. Switching to reduced homology thus gives short exact sequences

0 Ñ HnpXq Ñ HnpX,Aq Ñ Hn´1pAq Ñ 0 42 r r for n ě 1. By Hatcher, p. 125, HnpCiq » HnpX,Aq. By the previous argument, since the inclusion is null-homotopic, X must be a retract of Ci. By deﬁnition of retract, r ˝ i “ idX , so by functoriality r˚ ˝ i˚ “r pidX q˚, and thus the short exact sequence splits, yielding HnpX,Aq » HnpCiq » HnpXq ‘ Hn´1pAq for each n ě 1.

Exercise 94.r Let Xrbe path-connected,r locally path-connected, and semi-locally simply

connected. Let G1 Ă G2 be subgroups of π1pX, x0q. Let pi : XGi Ñ X be the covering

map corresponding to Gi. There is a covering space map f : XG1 Ñ XG2 such that p2 ˝ f “ p1.

Proof. The setup of this problem corresponds to Proposition 1.36 in Hatcher. In the

construction, XG1 and XG2 are quotients of the universal cover, and hence are path-connected covering spaces. Indeed, the correspondence refers to isomorphism classes of path-connected

covering spaces, Theorem 1.38 in Hatcher. Additionally, XG1 and XG2 must be locally path-connected as well, since a covering map is a local homeomorphism and X is locally path-connected. Now, consider the solid diagram

XG2 f p2 . X X G1 p1

1 ´1 2 ´1 Pick arbitrary basepoints x0 P p1 px0q and x0 P p2 px0q. Since p2 is a covering map and 1 XG1 is path-connected and locally path-connected, and pp1q˚pπ1pXG1 , x0qq “ G1 Ă G2 “ 2 pp2q˚pπ1pXG2 , x0qq by assumption, so by the lifting criterion, Proposition 1.33 in Hatcher, f exists making the diagram commute. It remains to show that f is a covering map. By assumption, p2 is a covering map,

so let tUxu denote the cover of X by evenly covered neighborhoods. For each x P XG2 , let Vx denote the neighborhood of x that maps homeomorphically onto the evenly covered ´1 neighborhood Ux of p2pxq. Since p2 ˝ f is a covering map, pp2 ˝ fq pUxq is a disjoint union

of open sets, or sheets, in XG1 . Note that, for any particular sheet Vx with p2pVxq “ Ux, ´1 ´1 ´1 ´1 pp2 ˝fq pp2pVxqq “ f pp2 pp2pVxqqq “ f pVxq is thus a disjoint union of open sets. Finally, ´1 since p2 ˝ f|f pVxq is a homeomorphism onto Ux, and p2|Vx is a homeomorphism onto Ux, ´1 f|f pVxq must be a homeomorphism onto Vx. Thus, f is a covering map, with tVxu providing

the cover of XG2 by evenly covered sets.

Exercise 95. Let X be the space obtained from the cube I3 by identifying opposite sides via the map translating the face by a unit distance in the normal direction and twisting by one-half of a full rotation. Below is a presentation of the fundamental group and all homology groups of X.

Proof. The cube I3 is homemorphic to the disk D3, so X is homeomorphic to D3 with antipodal points on BD3 identified, since the faces of the cube are rotated by π before identi- 3 fication. RP can be formed by identifying antipodes on S3. Furthermore, this identification can be restricted to identifying antipodes on the equator of S3, which is homeomorphic to 3 3 3 3 D . Hence, we have that X is homeomorphic to RP . First, π1pRP q » Z{2Z since S is 43 both a universal cover and a double cover. By Hatcher, Example 2.42, Z i “ 0 or i “ n odd. 3 HipRP q » $Z{2Z 0 ă i ă n, i odd &’0 otherwise A presentation for Z is given by xi ’y. That is, a single generator with no relations. A % 2 presentation for Z{2Z is given by Z{2Z » xi i “ 0y. Exercise 96. Let f : Y Ñ X be a fibration and let α : I Ñ X be a path from x to y. Apply the defining property of a fibration to the map f ´1pxq ˆ t0u Ñ f ´1pxq ˆ I ´1 ´1 to show that α induces a map α˚ : f pxq Ñ f pyq. The homotopy class of α˚ only depends on the homotopy class of α, and that in fact this construction yields a functor ΓX Ñ HTop, where ΓpXq is the fundamental groupoid of X. In particular, any two points in the same path-component of X have homotopy equivalent fibers.

´1 ´1 Proof. Construct from the path α a homotopy at : f pxq Ñ X given by atpf pxqq “ αptq. ´1 An initial lift a0 is given by the inclusion f pxq ãÑ Y . f is a ﬁbration, so by the homotopy ´1 lifting property, there exists a lift at : f pxq ˆ I Ñ Y . This construction yields the usual ﬁbration commutativer diagram

r a0 f ´1pxq Y

´1 r f . f pxqˆt0u a ´1 f pxq ˆ I ra X

´1 ´1 ´1 Note that a1 is a map α˚ : f pxq Ñ f pyq by commutativity since fpapf pxq, 1qq “ apf ´1pxq, 1q “ αp1q “ y. 1 : 1 Suppose rα » α rel BI via hps, tq I ˆ I Ñ X. We aim to show a˚ » a˚r. h gives maps ´1 ´1 1 hst : f pxq Ñ X given by hstpf pxqq “ hps, tq. Let h0,t “ a˚ and h1,t “ a˚. Let hs,0 be the ´1 ´1 inclusion f pxq ãÑ Y . These maps deﬁne an initial lift hBI : f pxq ˆ I ˆ BI Ñ Y . By Hatcher, p.405, f satisﬁes the homotopy liftingr property for anyr pair pZ ˆ I,Zr ˆ BIq. Thus there exists h such that r

´1 hBI r f pxq ˆ I ˆ BI Y r f h f ´1 x I I X p q ˆ ˆ rh

1 commutes. hs,1 then gives a homotopy a˚ to a˚ by commutativity. Hence, the homotopy class of each a˚ is independent of the choice of lift. It then follows that this construction is a well-definedr map F : ΓX Ñ HTop. Even better, F is actually a functor, as it preserves identity morphisms and composition of morphisms. It preserves the identity since a constant path αx at x gives a homotopy a˚ to a˚, which is the identity homotopy. It also preserves 1 1 composition since for the composition of two paths a ˚ a , the lift pa ˚ a q˚ is homotopy 1 1 equivalent to a˚a˚. To see this, if at is a lift where a1 “ a˚ and at is defined similarly, then 1 1 1 let at be defined as a2t for t P r0, 2 s and a2t´1 ˝ a˚ for t P r 2 , 1s. This gives a lift at where 1 a1 “ pa ˚ a q˚. r r r 44 r r r r r Finally, if α : I Ñ X is a path from x to y, then F pαq is a homotopy f ´1pxq to f ´1pyq with inverse F pα´1q, since id “ F pidq “ F pα´1 ˚ αq “ F pα´1q ˝ F pαq by functoriality. Thus 1 1 a path x to y gives a homotopy equivalence f ´ pxq to f ´ pyq.

Exercise 97. Let X be a space. HnpX; Zq “ 0 for all n ą 0 if, and only if, HnpX; Qq “ 0 and HnpX; Z{pZq “ 0 for all prime numbers p and all n ą 0. Proof. By the universal coeﬃcient theorem, the sequence

0 Ñ HipX; Zq b A Ñ HipX; Aq Ñ TorpHi´1pX; Zq,Aq Ñ 0 is exact for any abelian group A. p ùñ q If HnpX; Zq “ 0 for all n ą 0, the sequence above becomes

0 Ñ HnpX; Aq Ñ TorpHn´1pX; Zq,Aq Ñ 0 for all n ą 0, since 0 b A “ 0 as any 0 b a is the zero element by Hatcher, p. 215. Thus, by exactness, HnpX; Aq » TorpHn´1pX; Zq,Aq. For n ą 1, TorpHn´1pX; Zq,Aq “ 0 since Hn´1pX; Zq “ 0 and 0 is torsion-free. Additionally, for n “ 1, H0pX; Zq counts the path-components of X, so it is isomorphic to Zn for some n, which is also torsion-free. Hence, TorpHn´1pX; Zq,Aq “ 0 for all n ą 0 by Proposition 3A.5 in Hatcher and thus HnpX; Aq “ 0 for n ą 0 for both A “ Q and A “ Z{pZ for all prime p. p ðù q If HnpX, Qq “ 0 for all n ą 0, setting A “ Q, the sequence above yields that

0 Ñ HnpX; Zq b Q Ñ 0 Ñ TorpHi´1pX; Zq, Qq Ñ 0 is exact, so by exactness, HnpX; Zq b Q “ 0. We claim HnpX; Zq » HnpX; Zq b Q, which then ﬁnishes the proof as HnpX; Zq b Q “ 0. HnpX; Z{pZq “ 0 for all n ą 0 and all primes p if and only if HnpX; Zq is a vector space over Q for all n ą 0. Also, by Hatcher, p. 215, HnpXq b Q “ HnpXq bQ Q. This reduces what is left to showing that for X a vector space over Q, X bQ Q » X. But it is a general algebraic fact that if M is an R-module, then M bR R » M.

Exercise 98. If X and Y are pointed spaces and n ě 2,

πnpX _ Y q » πnpXq ‘ πnpY q ‘ πn`1pX ˆ Y,X _ Y q.

Proof. Consider the long exact sequence of relative homotopy groups

B i˚ ... Ñ πn`1pX ˆ Y,X _ Y q ÝÑ πnpX _ Y q ÝÑ πnpX ˆ Y q Ñ ... where i˚ is the induced map of the inclusion i : X _ Y Ñ X ˆ Y . We claim that there is a right splitting map j˚ such that i˚ ˝ j˚ “ id˚. This claim gives that πnpX _ Y q » πnpX ˆ Y q ‘ πn`1pX ˆ Y,X _ Y q, from which the proposition follows by Proposition 4.2 in Hatcher, which gives πnpX ˆ Y q » πnpXq ˆ πnpY q. Now to address the claim. Consider any element rγs P πnpX ˆ Y q, represented by γ : n n S Ñ X ˆ Y . Composing with the natural projections px and py gives px ˝ γ : S Ñ X and n n py ˝ γ : S Ñ Y , which when composed with the inclusion give i ˝ px ˝ γ : S Ñ X ãÑ X _ Y n x y and i ˝ py ˝ γ : S Ñ Y ãÑ X _ Y . Re-label these γ and γ respectively. Now consider the x y x y equivalence class rγ ˚ γ s P πnpX _ Y q. Deﬁne a map j˚prγsq “ rγ ˚ γ s. It remains to show that j˚ is a homomorphism and that i˚ ˝ j˚ “ id˚. 45 The claim that j˚ is a homomorphism will require that πn is abelian, so consider n ě 2. x x x Note that the projection and inclusions induce homomorphisms, so pγ1 ˚ γ2q “ γ1 ˚ γ2 , and similarly for y. Explicitly computing,

j˚prγ1srγ2sq “ j˚prγ1 ˚ γ2sq x y “ rpγ1 ˚ γ2q ˚ pγ1 ˚ γ2q s x x y y “ rγ1 ˚ γ2 ˚ γ1 ˚ γ2 s x y x y “ rγ1 ˚ γ1 ˚ γ2 ˚ γ2 s x y x y “ rγ1 ˚ γ1 srγ2 ˚ γ2 s

“ j˚prγ1sqj˚prγ2sq x y Now, to prove i˚˝j˚ “ id˚, we show a homotopy ipγ1 ˚γ1 q » γ, which gives that i˚pj˚prγsq “ x y x y x y x y i˚prγ1 ˚γ1 sq “ rγs. Rewriting, γ “ ppxpγ1 q, pypγ1 qq, and ipγ1 ˚γ1 q “ ppxpγ1 q, y0q˚px0, pypγ1 qq, x y so the explicit homotopy γ ÞÑ γ1 ˚ γ1 is given by the usual reparametrizing homotopy used to show that πn is a group.